Your inequality is correct. Below is my proof, which took me only eight hours, and which shows that the inequality can be improved.

For the proof, I note that, since abc = 1, we have $ \sqrt [3]{abc}\equal{} 1$. On the other hand, by the GM-HM inequality, we have $ \sqrt [3]{abc}\geq\frac{3}{1/a\plus{}1/b\plus{}1/c}$, and thus $ 1\geq\frac{3}{1/a\plus{}1/b\plus{}1/c}$. This yields

$ \frac{b\plus{}c}{a^{3}\plus{}bc}\leq\frac{b\plus{}c}{a^{3}\plus{}bc\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}$.

Similarly,

$ \frac{c\plus{}a}{b^{3}\plus{}ca}\leq\frac{c\plus{}a}{b^{3}\plus{}ca\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}$

and

$ \frac{a\plus{}b}{c^{3}\plus{}ab}\leq\frac{a\plus{}b}{c^{3}\plus{}ab\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}$.

Now we will show:

[b]Lemma 1.[/b] For any three positive numbers a, b, c, we have

$ \frac{b\plus{}c}{a^{3}\plus{}bc\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\plus{}\frac{c\plus{}a}{b^{3}\plus{}ca\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\plus{}\frac{a\plus{}b}{c^{3}\plus{}ab\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}$
$ \leq\frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}}$.

Of course, this Lemma 1 will then yield the problem, since we will have

$ \frac{b\plus{}c}{a^{3}\plus{}bc}\plus{}\frac{c\plus{}a}{b^{3}\plus{}ca}\plus{}\frac{a\plus{}b}{c^{3}\plus{}ab}$
$ \leq\frac{b\plus{}c}{a^{3}\plus{}bc\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\plus{}\frac{c\plus{}a}{b^{3}\plus{}ca\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\plus{}\frac{a\plus{}b}{c^{3}\plus{}ab\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}$
$ \leq\frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}}$.

Hence, in order to solve the problem, it is enough to prove Lemma 1.

[i]Proof of Lemma 1.[/i] Denoting $ x \equal{}\frac{1}{a}$, $ y \equal{}\frac{1}{b}$, $ z \equal{}\frac{1}{c}$, we have $ a \equal{}\frac{1}{x}$, $ b \equal{}\frac{1}{y}$, $ c \equal{}\frac{1}{z}$. Thus, after some algebra,

$ \frac{b\plus{}c}{a^{3}\plus{}bc\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\equal{}\frac{x^{3}\left(y\plus{}z\right)\left(x\plus{}y\plus{}z\right)}{2x^{3}\plus{}\left(x^{3}\plus{}yz^{2}\plus{}zy^{2}\plus{}xyz\right)}$.

By the AM-GM inequality, $ x^{3}\plus{}yz^{2}\plus{}zy^{2}\plus{}xyz\geq 4\sqrt [4]{x^{3}\cdot yz^{2}\cdot zy^{2}\cdot xyz}\equal{} 4\sqrt [4]{x^{4}y^{4}z^{4}}\equal{} 4xyz$, and thus,

$ \frac{b\plus{}c}{a^{3}\plus{}bc\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\equal{}\frac{x^{3}\left(y\plus{}z\right)\left(x\plus{}y\plus{}z\right)}{2x^{3}\plus{}\left(x^{3}\plus{}yz^{2}\plus{}zy^{2}\plus{}xyz\right)}$
$ \leq\frac{x^{3}\left(y\plus{}z\right)\left(x\plus{}y\plus{}z\right)}{2x^{3}\plus{}4xyz}\equal{}\frac{x\plus{}y\plus{}z}{2}\cdot\frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}$.

Similarly,

$ \frac{c\plus{}a}{b^{3}\plus{}ca\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\equal{}\frac{x\plus{}y\plus{}z}{2}\cdot\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}$

and

$ \frac{a\plus{}b}{c^{3}\plus{}ab\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\equal{}\frac{x\plus{}y\plus{}z}{2}\cdot\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}$.

Now we come to another lemma:

[b]Lemma 2.[/b] For any three positive reals x, y, z, the inequality

$ \frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\plus{}\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\plus{}\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}\leq\frac{2\left(x^{2}\plus{}y^{2}\plus{}z^{2}\right)}{x\plus{}y\plus{}z}$

holds.

From this lemma, we will be able to conclude

$ \frac{x\plus{}y\plus{}z}{2}\cdot\left(\frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\plus{}\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\plus{}\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}\right)\leq x^{2}\plus{}y^{2}\plus{}z^{2}$;

thus, using what we have found before,

$ \frac{b\plus{}c}{a^{3}\plus{}bc\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\plus{}\frac{c\plus{}a}{b^{3}\plus{}ca\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}\plus{}\frac{a\plus{}b}{c^{3}\plus{}ab\cdot\frac{3}{1/a\plus{}1/b\plus{}1/c}}$
$ \leq\frac{x\plus{}y\plus{}z}{2}\cdot\frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\plus{}\frac{x\plus{}y\plus{}z}{2}\cdot\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\plus{}\frac{x\plus{}y\plus{}z}{2}\cdot\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}$
$ \equal{}\frac{x\plus{}y\plus{}z}{2}\cdot\left(\frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\plus{}\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\plus{}\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}\right)$
$ \leq x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}\frac{1}{a^{2}}\plus{}\frac{1}{b^{2}}\plus{}\frac{1}{c^{2}}$

(since $ x \equal{}\frac{1}{a}$, $ y \equal{}\frac{1}{b}$, $ z \equal{}\frac{1}{c}$); thus, Lemma 1 will be proven. Hence, it only remains to prove Lemma 2.

[i]Proof of Lemma 2.[/i] We note that

$ \frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\equal{}\left(y\plus{}z\right)\minus{}\frac{2\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}$.

Similarly,

$ \frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\equal{}\left(z\plus{}x\right)\minus{}\frac{2\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}$.
$ \frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}\equal{}\left(x\plus{}y\right)\minus{}\frac{2\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}$.

Hence,

$ \frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\plus{}\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\plus{}\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}$
$ \equal{}\left(\left(y\plus{}z\right)\minus{}\frac{2\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}\right)\plus{}\left(\left(z\plus{}x\right)\minus{}\frac{2\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}\right)\plus{}\left(\left(x\plus{}y\right)\minus{}\frac{2\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}\right)$
$ \equal{} 2\left(\left(x\plus{}y\plus{}z\right)\minus{}\left(\frac{\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}\plus{}\frac{\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}\plus{}\frac{\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}\right)\right)$.

Therefore, the inequality that we have to prove,

$ \frac{x^{2}\left(y\plus{}z\right)}{x^{2}\plus{}2yz}\plus{}\frac{y^{2}\left(z\plus{}x\right)}{y^{2}\plus{}2zx}\plus{}\frac{z^{2}\left(x\plus{}y\right)}{z^{2}\plus{}2xy}\leq\frac{2\left(x^{2}\plus{}y^{2}\plus{}z^{2}\right)}{x\plus{}y\plus{}z}$,

is equivalent to

$ 2\left(\left(x\plus{}y\plus{}z\right)\minus{}\left(\frac{\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}\plus{}\frac{\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}\plus{}\frac{\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}\right)\right)\leq\frac{2\left(x^{2}\plus{}y^{2}\plus{}z^{2}\right)}{x\plus{}y\plus{}z}$,

i. e. to

$ \left(x\plus{}y\plus{}z\right)\minus{}\left(\frac{\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}\plus{}\frac{\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}\plus{}\frac{\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}\right)\leq\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{x\plus{}y\plus{}z}$.

Subtracting both sides from x + y + z, we get

$ \frac{\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}\plus{}\frac{\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}\plus{}\frac{\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}\geq\left(x\plus{}y\plus{}z\right)\minus{}\frac{x^{2}\plus{}y^{2}\plus{}z^{2}}{x\plus{}y\plus{}z}$,

what simplifies to

$ \frac{\left(y\plus{}z\right)yz}{x^{2}\plus{}2yz}\plus{}\frac{\left(z\plus{}x\right)zx}{y^{2}\plus{}2zx}\plus{}\frac{\left(x\plus{}y\right)xy}{z^{2}\plus{}2xy}\geq\frac{2\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$.

Now, a rather strange way to rewrite this is to make the fractions on the left hand side more complicated:

$ \frac{\left(\left(y\plus{}z\right)yz\right)^{2}}{\left(x^{2}\plus{}2yz\right)\left(y\plus{}z\right)yz}\plus{}\frac{\left(\left(z\plus{}x\right)zx\right)^{2}}{\left(y^{2}\plus{}2zx\right)\left(z\plus{}x\right)zx}\plus{}\frac{\left(\left(x\plus{}y\right)xy\right)^{2}}{\left(z^{2}\plus{}2xy\right)\left(x\plus{}y\right)xy}\geq\frac{2\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$.

Now, applying [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in the Engel form[/url], we have

$ \frac{\left(\left(y\plus{}z\right)yz\right)^{2}}{\left(x^{2}\plus{}2yz\right)\left(y\plus{}z\right)yz}\plus{}\frac{\left(\left(z\plus{}x\right)zx\right)^{2}}{\left(y^{2}\plus{}2zx\right)\left(z\plus{}x\right)zx}\plus{}\frac{\left(\left(x\plus{}y\right)xy\right)^{2}}{\left(z^{2}\plus{}2xy\right)\left(x\plus{}y\right)xy}$
$ \geq\frac{\left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}}{\left(x^{2}\plus{}2yz\right)\left(y\plus{}z\right)yz\plus{}\left(y^{2}\plus{}2zx\right)\left(z\plus{}x\right)zx\plus{}\left(z^{2}\plus{}2xy\right)\left(x\plus{}y\right)xy}$
$ \equal{}\frac{\left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}}{2\left(x\plus{}y\plus{}z\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)}$.

Now, the last important lemma in our proof will be:

[b]Lemma 3.[/b] For any three positive reals x, y, z, we have

$ \left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}\geq 4\left(yz\plus{}zx\plus{}xy\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)$.

In fact, once this Lemma 3 will be shown, we will have

$ \frac{\left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}}{2\left(x\plus{}y\plus{}z\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)}\geq\frac{4\left(yz\plus{}zx\plus{}xy\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)}{2\left(x\plus{}y\plus{}z\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)}$
$ \equal{}\frac{2\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$,

and thus we will have

$ \frac{\left(\left(y\plus{}z\right)yz\right)^{2}}{\left(x^{2}\plus{}2yz\right)\left(y\plus{}z\right)yz}\plus{}\frac{\left(\left(z\plus{}x\right)zx\right)^{2}}{\left(y^{2}\plus{}2zx\right)\left(z\plus{}x\right)zx}\plus{}\frac{\left(\left(x\plus{}y\right)xy\right)^{2}}{\left(z^{2}\plus{}2xy\right)\left(x\plus{}y\right)xy}$
$ \geq\frac{\left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}}{2\left(x\plus{}y\plus{}z\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)}\geq\frac{2\left(yz\plus{}zx\plus{}xy\right)}{x\plus{}y\plus{}z}$,

what will finish the proof of Lemma 2. Thus, all we need is to show Lemma 3.

[i]First proof of Lemma 3.[/i] We have

$ \left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}\minus{}4\left(yz\plus{}zx\plus{}xy\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)$
$ \equal{} y^{2}z^{2}\left(y\minus{}z\right)^{2}\plus{}z^{2}x^{2}\left(z\minus{}x\right)^{2}\plus{}x^{2}y^{2}\left(x\minus{}y\right)^{2}$
$ \plus{}2xyz\left(x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}3xyz\minus{}yz^{2}\minus{}zy^{2}\minus{}zx^{2}\minus{}xz^{2}\minus{}xy^{2}\minus{}yx^{2}\right)$,

what is $ \geq 0$ since squares are nonnegative and since

$ x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}3xyz\minus{}yz^{2}\minus{}zy^{2}\minus{}zx^{2}\minus{}xz^{2}\minus{}xy^{2}\minus{}yx^{2}\geq 0$

by Schur. Hence,

$ \left(\left(y\plus{}z\right)yz\plus{}\left(z\plus{}x\right)zx\plus{}\left(x\plus{}y\right)xy\right)^{2}\geq 4\left(yz\plus{}zx\plus{}xy\right)\left(y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)$,

and Lemma 3 is proven.

[i]Second proof of Lemma 3.[/i] We have

$ \left(\left( y\plus{}z\right) yz\plus{}\left( z\plus{}x\right) zx\plus{}\left( x\plus{}y\right)xy\right)^{2}\equal{}\left(\left( x\cdot zx\plus{}y\cdot xy\plus{}z\cdot yz\right)\plus{}\left( x\cdot xy\plus{}y\cdot yz\plus{}z\cdot zx\right)\right)^{2}$
$ \geq4\left( x\cdot zx\plus{}y\cdot xy\plus{}z\cdot yz\right)\left( x\cdot xy\plus{}y\cdot yz\plus{}z\cdot zx\right)$

(by the inequality $ \left( u\plus{}v\right)^{2}\geq4uv$ which holds for any two reals $ u$ and $ v$). Hence, it remains to prove that

$ \left( x\cdot zx\plus{}y\cdot xy\plus{}z\cdot yz\right)\left( x\cdot xy\plus{}y\cdot yz\plus{}z\cdot zx\right)\geq\left( yz\plus{}zx\plus{}xy\right)\left( y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)$.

Since this inequality is symmetric in its variables $ x$, $ y$, $ z$, we can WLOG assume that $ x\geq y\geq z$.

Denote $ A\equal{}x$, $ B\equal{}y$, $ C\equal{}z$. Then, $ A\geq B\geq C$. Also, denote $ X\equal{}yz$, $ Y\equal{}zx$, $ Z\equal{}xy$. Then, $ y\leq x$ yields $ yz\leq zx$, so that $ X\leq Y$. Hence, applying Theorem 9 [b]a)[/b] from [url=http://www.mathlinks.ro/viewtopic.php?mode=attach&id=10141]my article "The Vornicu-Schur inequality and its variations"[/url] to the three reals $ A$, $ B$, $ C$ and the three nonnegative reals $ X$, $ Y$, $ Z$, we get

$ \left( AY\plus{}BZ\plus{}CX\right)\left( AZ\plus{}BX\plus{}CY\right)\geq\left( X\plus{}Y\plus{}Z\right)\left( XBC\plus{}YCA\plus{}ZAB\right)$.

This rewrites as

$ \left( x\cdot zx\plus{}y\cdot xy\plus{}z\cdot yz\right)\left( x\cdot xy\plus{}y\cdot yz\plus{}z\cdot zx\right)\geq\left( yz\plus{}zx\plus{}xy\right)\left( yz\cdot yz\plus{}zx\cdot zx\plus{}xy\cdot xy\right)$.

Equivalently,

$ \left( x\cdot zx\plus{}y\cdot xy\plus{}z\cdot yz\right)\left( x\cdot xy\plus{}y\cdot yz\plus{}z\cdot zx\right)\geq\left( yz\plus{}zx\plus{}xy\right)\left( y^{2}z^{2}\plus{}z^{2}x^{2}\plus{}x^{2}y^{2}\right)$,

and Lemma 3 is proven.

$ \blacksquare$

  Darij

Congratulations, Darij, for your very nice and very complete proof!!

wow ,nice proof ,long proof , you must have taken a long time to post this proof
          :D

[quote="pascual2004"]Congratulations for the longest post ever :D  :D[/quote]

Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15620]here.[/url] The second posting by Peter DVD is not too short as well.  :lol:


WLOG, $c = \min \{a,b,c \}$. Then $ax+by+cz = (a-c)x+(b-c)y+c \geq c$.
 Since $ax+by+cz \rightarrow c$ for $z \rightarrow 1$,  $\inf \{ax+by+cz | x,y,z>0; \ x+y+z=1 \}= c$.
It is possible to achieve this infimum if and only if $(a-c)=(b-c)=0 \Leftrightarrow a=b=c$

Follow you it only has $inf$ when $a=b=c$ but $a,b,c$ is const!! when $a\not=b\not=c$ then Does it have $inf$ I think here we need find condition of variable $x,y,z$ not $a,b,c$.
Maybe it has $min =\frac{1}{3}(a+b+c)$ when $x=y=z=\frac{1}{3}$???
am I true? :maybe:

The infimum is achieved when $x=y=0,z=1,$ assuming $c=\min\{a,b,c\}.$ The supremum is achieved when $x=1,y=z=0.$

[quote="gemath"]Follow you it only has $inf$ when $a=b=c$ but $a,b,c$ is const!! when $a\not=b\not=c$ then Does it have $inf$ I think here we need find condition of variable $x,y,z$ not $a,b,c$. [/quote]

I meant it has a [i]minimum[/i] if and only if $a=b=c$. $\min \{ a,b,c \}$ is always the [i]infimum[/i] of $ax+by+cz$.

You misunderstood me, gemath. Maybe I have trouble with the technical vocabulary.

[quote="gemath"]Given $a,b,c>0$ is const and $x,y,z>0,x+y+z=1$ find min value of
$ax+by+cz$
Is there a solution for this problem? Please help me! :)[/quote]
Problem 4 from 1st round of LVI Polish MO (very similar to this):
[i][color=blue]Given constant $a,b\in\mathbb{R}^{+}$ and $n\in\mathbb{N}^{+}$. Find the maximum of $x_{1}y_{1}+\ldots+x_{n}y_{n}$,
where $x_{1},\ldots,x_{n}$ and $y_{1},\ldots,y_{m}$ belong to interval $\langle 0,1\rangle$ and satisfy $x_{1}+\ldots+x_{n}\leq a$ and $y_{1}+\ldots+y_{n}\leq b$. [/color][/i]


A 4" x 6.5" picture is to be mounted on poster-board so that there is a border of uniform width. Find the width of the border in inches if the area of the border is 20 sq. in.

Answer:
[hide]
0.82
[/hide]

Let $ x$ equal the width of the border.  We have $ (6.5\plus{}2x)(4\plus{}2x)\minus{}4*6.5\equal{}20\Rightarrow 4x^2\plus{}21x\minus{}20\equal{}0$.  Solving, we find $ x\equal{}\dfrac{\minus{}21\pm
\sqrt{761}}{8}$.  We take the positive root, which is $ \approx 0.82$

i did the same thing, but i didn't use 2x. Why'd you use 2x?

Because the border is on two sides, not just one.


if I'm writing a document and i want to number an equation but I don't like the default numbers, is there a way to make my own equation numbers?

Soemthing like this:

\begin{equation*} 1+1\neq 3 \tag{$73$}\end{equation*}

(I'm not sure if the *s are necessary.) If you don't want the parentheses around the equation number/symbol/whatever, you can use \tag* instead of \tag.


i have recently been learning induction so i was wondering if this would be a completely correct proof
prove that $n^{3}-n$ for any integer $n$ is divisible by 3
starting by
$n=0$, $0^{3}-0=0$ which is obviously divisible by 3
so if $S_{k}$ holds, then must show that $S_{k+1}$ and $S_{k-1}$ holds
so assuming that $n^{3} - n \mod{3} \equiv 0$
then we can add $3n^2+3n$ to the previous because we know that
$3n^2+3n \mod{3}$
$3(n^2+n) \mod{3}$
$3 \mod{3} \equiv 0$ so
$(0)(n^2+n) \mod{3} \equiv 0$ so
$n^{3} - n + 3n^2 + 3n \mod{3} \equiv 0$
$n^{3} + 3n^2 + 3n + 1 - n - 1 \mod{3} \equiv 0$
$(n +1)^{3} - (n + 1) \mod{3} \equiv 0$
therefore $S_{k}$ implies that $S_{k+1}$ is true and all positive integers are true
now we have to show that $S_{k+1}$ implies $S_{k}$
or that $S_{k}$ implies $S_{k-1}$
so similarily to before, we assume that
$n^{3} - n \mod{3} \equiv 0$
we know that $-3n^2 +3n \mod{3} \equiv 0$
since $3 \mod{3} \equiv 0$
so $n^{3} -3n^2 +3n - n \mod{3} \equiv 0$
$(n^{3} -3n^2 +3n - 1) - n + 1 \mod{3} \equiv 0$ or
$(n -1)^{3} - (n - 1) \mod{3} \equiv 0$
therefore  $S_{k}$ implies $S_{k-1}$ and all negative integers are proven
since the statement is proven for negative integers, positive integers, and 0, therefore the statement is true for all integers n

I didn't read your solution :blush:
But to Answer your question.It is so easy.$n^3-1=(n-1)n(n+1)$ so one of them is divisible by $3$. ;)

I think you should show for the case $n=1$ and $n=2$ also since $0$ divisible by $3$ sounds like showing nothing  :?

i know that it is obviously true, but that is not proving through induction, and my purpose is to practice induction, this is a problem that i have done before, but just saying it is obvious does not really prove anything, what i did before was that i said that you could say that any integer can be written as either $3m$, $3m +1$, or $3m+2$ (this can be shown by mods) then placing those into $n^{3}-n$ always gives something that is divisible by 3, but as i said before, this is not induction

and for showing the case of 0, that seems that is correct because then that implies the cases for 2,3,etc. so those cases are implied

and this problem leads me to another problem, that i am not sure to prove but it is obvious, but if you are given something like this, you probably have to use induction

Prove that: $\prod^{n}_{k=1}(x-k) \mod{n!} \equiv 0$ for any integer $n>0$ and any integer $x$


inequalities
function
calculus
IMO Shortlist

Let $n$ be a positive integer and let $(x_1,\ldots,x_n)$, $(y_1,\ldots,y_n)$ be two sequences of positive real numbers.  Suppose $(z_2,\ldots,z_{2n})$ is a sequence of positive real numbers such that $z_{i+j}^2 \geq x_iy_j$ for all $1\le i,j \leq n$.

Let $M=\max\{z_2,\ldots,z_{2n}\}$.  Prove that \[
	\left( \frac{M+z_2+\dots+z_{2n}}{2n} \right)^2
	\ge
	\left( \frac{x_1+\dots+x_n}{n} \right)
	\left( \frac{y_1+\dots+y_n}{n} \right). \]

[hide="comment"]
[i]Edited by Orl.[/i]
[/hide]

[i]Proposed by Reid Barton, USA[/i]

I'm curious who can share with us a solution. This is the most unconventional and hard inequality I have ever seen. Absolutely impossible. As far as I heard, no contestant in Mathlinks Contest managed to do it. Is there anyone who can supply a proof?

I have thought on this proble nearly 1 hour and I obtained nothing.
It's impossible.  :(  :(  :(
Do you know which country sent this problem. :?  :?  :?  :(  :(  :(

[quote="harazi"]Ha, ha, I like that with one hour. So, should I understand that every inequality except this one, you managed to prove it in one hour? Boy, you really are tough! I spent two days on this problem and couldn't find anything. But I'm not the only one, since I heard that none from the leaders in OIM could solve it. This problem is created by Reid Barton. I think it says something, right?[/quote]

Who is Reid Barton?

One of the best IMO contestants ever - he scored four gold medals I think, and got full marks at IMO 2001.

I have proved following inequality:
\[ S^2\geq (x_1+x_2+\dots x_n)(y_1+y_2+\dots+y_n),\]
where $S=\max\limits_{i=1,\dots,n}(z_{i+1}+z_{i+1}+\dots z_{i+n})$.

Did anybody prove other similar inequalities?

I think this inequality is a kind of improvement of classical Cotlar-Stein inequality:

Lemma (Cotlar, Stein).
Let $z_i\geq 0$, $i\in\mathbb{Z}$, and $S=\sum\limits_{-\infty}^{+\infty}z_i<+\infty$.
Let $T_1,\,T_2,\,\dots,\,T_n$ be linear continuous operators in Hilbert space $H$, $||T_i^*T_j||\leq z_{i-j}^2$ and $||T_iT_j^*||\leq z_{i-j}^2$ $\forall\,i,j\in\{1,2,\dots,n\}$.
Then $||T_1+T_2+\dots+T_n||\leq S$.

Unfortunately in our case using of this inequality gives too bad estimation, but its technique brings us to inequality from my previous post.

The official solution is this (im just copying from the booklet :) ): and its really nice:

Let X = max (x1,x2 ... x(n)), and Y = max (y1,y2 ... y(n)). By replacing with
x(i)'=x(i)/X, and y(i)'=y(i)/Y and z(i) by z(i)'=z(i)/sqrt XY, we may assume
X=Y=1. Now we prove the following:
M + z2 + ... + z(2n) >= x1 + x2 ... + x(n) + y1 + y2 ... + y(n) ... (*)
from which the result follows with am-gm.
to prove (*) we will show that for any r>=0, the number of terms greater than r on the left hand side is at leas the number of such terms on the right-hand side. Then the k-th largest term on the left hand side is greater than or equal to the number of such terms on the right hand side for each k proving (*) (really wierd ..).
If r>=1 then there are no terms greater than r on the right-hand side. So suppose r<1. Let A = {1<=i<=n|x(i) > r}, a = |A|, B = {1<=i<=n|y(i) > r}, b = |B|.
Since max { x1, x2, ... x(n)} = max {y1, y2 ... y(n)} = 1, both a and b are at least 1. Now x(i) > r and y(j) > r implies z(i+j) >= sqrt (x(i)*y(j)) > r,
so C = {2 <= i <= 2n | z(i) > r} is at least as large as
A+B = {x+y | x in A, y in B}.
However we know |A+B| >= |A| + |B| - 1 because if A={i1,i2 .. i(a)},
i1<i2 .. <i(a) and B ={j1,j2 .. j(b)}, j1<j2 ... <j(b) then the a+b-1 numbers
i1+j1, i1+j2 ... i1+jb, i2+jb, ... i(a) + j(b) are all distinct and belong to A+B.
Hence |C| >= a+b-1. In particular |C| >= 1 so z(k) > r for some k. Then M>r, so the left hand side of (*) has at least a+b terms greater than r. Since a+b is the number of terms greater than r on the right hand side we have proven (*).

[quote]to prove (*) we will show that for any r>=0, the number of terms greater than r on the left hand side is at least the number of such terms on the right-hand side. Then the k-th largest term on the left hand side is greater than or equal to the number of such terms on the right hand side for each k proving (*) (really wierd ..).
[/quote]

I really don't understand this step.

Has anyone studied and understood it ?

Thank you very much.

Hi Manlio!

[quote="manlio"][quote]to prove (*) we will show that for any r>=0, the number of terms greater than r on the left hand side is at least the number of such terms on the right-hand side. Then the k-th largest term on the left hand side is greater than or equal to the number of such terms on the right hand side for each k proving (*) (really wierd ..).
[/quote]

I really don't understand this step.[/quote]

I'm sorry for the belated reply, I also had my troubles with this passage. In fact, it is a really ingenious argument but I don't understand how Vinoth_90_2004 justifies it. Just let me tell you how [i]I[/i] justify it:

The crux here is the following lemma:

[b]Lemma 1.[/b] Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be positive real numbers. Assume that for every positive real number r, the number of all i such that $a_i > r$ is greater or equal to the number of all i such that $b_i > r$. Then, $a_1 + a_2 + ... + a_n \geq b_1 + b_2 + ... + b_n$.

[i]Proof.[/i] For any positive real number r, let the A(r) denote the number of all i such that $a_i > r$, and let B(r) denote the number of all i such that $b_i > r$. Then, we know that $A\left( r\right) \geq B\left( r\right)$ holds for all $r \geq 0$. Moreover, the functions A(r) and B(r) are piecewisely constant functions, and hence we can integrate them. (Actually, I will integrate them out of laziness. The proof can be rewritten in a completely elementary form, since instead of integrating piecewisely constant functions, you can just write down a sum of finitely many rectangle areas.)

Remember that the numbers $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ are all positive, and we can interchange the $a_1$, $a_2$, ..., $a_n$ arbitrarily and interchange the $b_1$, $b_2$, ..., $b_n$ arbitrarily without changing everything in the assertion of the theorem. Hence, we can WLOG assume that $0 < a_1 \leq a_2 \leq ... \leq a_n$ and $0 < b_1 \leq b_2 \leq ... \leq b_n$. Finally, there obviously exists a positive number q which is greater than all of the numbers $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$. Then, we can easily see that

  [tex]\int\limits_0^q A\left( r\right) dr = a_1 + a_2 + ... + a_n[/tex]

and

  [tex]\int\limits_0^q B\left( r\right) dr = b_1 + b_2 + ... + b_n[/tex].

Now, since we have $A\left( r\right) \geq B\left( r\right)$ for all $r \geq 0$, we also get

  [tex]\int\limits_0^q A\left( r\right) dr \geq \int\limits_0^q B\left( r\right) dr [/tex],

so that $a_1 + a_2 + ... + a_n \geq b_1 + b_2 + ... + b_n$. This proves Lemma 1.

Oh yes, I really wonder how Vinoth_90_2004's proof (or, more exactly, the proof from the Shortlist booklet) goes. I hope it is simpler than mine...

  Darij

[quote="darij grinberg"][b]Lemma 1.[/b] Let $a_1$, $a_2$, ..., $a_n$, $b_1$, $b_2$, ..., $b_n$ be positive real numbers. Assume that for every positive real number r, the number of all i such that $a_i > r$ is greater or equal to the number of all i such that $b_i > r$. Then, $a_1 + a_2 + ... + a_n \geq b_1 + b_2 + ... + b_n$.

[...]

  [tex]\int\limits_0^q A\left( r\right) dr \geq \int\limits_0^q B\left( r\right) dr [/tex],

Oh yes, I really wonder how Vinoth_90_2004's proof (or, more exactly, the proof from the Shortlist booklet) goes. I hope it is simpler than mine...

  Darij[/quote]

Very nice!

In a private message to manlio, I actually proved that $a_k\geq b_k$ if both sequences are sorted in increasing order. Is it possible to explain this using integrals?

PS. I can't copy my message here because I can't see the quote in the Sentbox and I don't want to re-type it, maybe manlio can see the quote from his Inbox.

[quote="fuzzylogic"]I actually proved that $a_k\geq b_k$ if both sequences are sorted in increasing order.[/quote]

Oh yes. This was the cook!

I must have been too blind to note this simple argument.

Especially if we use my notations A(r) and B(r) and my WLOG assumption $0 < a_1 \leq a_2 \leq ... \leq a_n$ and $0 < b_1 \leq b_2 \leq ... \leq b_n$, this argument becomes trivial: For every k, we have $A\left( a_k \right) \leq n-k$. This is because $A\left( a_k \right) $ is the number of all i such that $a_i > a_k$, and actually there are at most n - k such numbers i (in fact, in the usual case, there are [i]exactly[/i] n - k such numbers i, but if $a_k = a_{k+1}$, this number becomes smaller). Now we have by assumption $A\left( r\right) \geq B\left( r\right)$ for every positive r, hence we have particularly $A\left( a_k \right) \geq B\left( a_k \right) $. Together with $A\left( a_k \right) \leq n-k$, this yields $B\left( a_k\right) \leq n-k$. Now, if we would have $a_k < b_k$, then we also would have $a_k < b_k \leq b_{k+1} \leq ... \leq b_n$, so we would already have at least n - k + 1 numbers i such that $b_i > a_k$. Hence, the value $B\left( a_k\right) $ would be at least n - k + 1. But this contradicts $B\left( a_k\right) \leq n-k$. Hence we can never have $a_k < b_k$. Thus, we must have $a_k \geq b_k$ for every k. Summing up, this yields $a_1 + a_2 + ... + a_n \geq b_1 + b_2 + ... + b_n$. This proves Lemma 1.

Yes, Fuzzylogic, this is not only a proof which is much simpler than mine; now I also understand that this is exactly the same argument Vinoth_90_2004 gave! But after all, what I did was a pretty neat exercise on applying integral calculus, wasn't it? :?

  Darij

This inequality is a particular case of Prekopa-Leindler inequality. It says that if we take the function $ h(t) = \sup \{f(x)g(y) \mid x+y=t\} $ then $\int_0^{\infty} h(t)dt$ is at least $ 2\sqrt{\int_0^{\infty} f^2(x) dx} \cdot \sqrt{\int_0^{\infty} g^2(x)dx} $. This explains the official solution.

This is the proof of Fuzzylogic.

[quote="fuzzylogic"][quote]to prove (*) we will show that for any r>=0, the number of terms greater than r on the left hand side is at least the number of such terms on the right-hand side. Then the k-th largest term on the left hand side is greater than or equal to the number of such terms on the right hand side for each k proving (*) (really wierd ..).
[/quote]

I think I understand the logic. Let me try to rephrase it as a lemma as follows:

[b]Lemma[/b] Let $0< a_1\le a_2 \le \cdots \le a_m$ and $0 < b_1 \le b_2 \le \cdots \le b_m$ be two increasingly sorted sequences of positive reals.

For any real $r$, define $A_r = \{i \,|\, a_i > r\}$ and $B_r = \{i \,|\, b_i > r\}$.

Assume that $|A_r| \geq |B_r|$ for any $r\ge 0$, then $a_k \ge b_k$ for any $1\le k \le m$.

[b]Proof[/b]: (by contradiction) Suppose $a_k < b_k$ for some $1\le k \le m$. Pick any real $r$ such that $a_k<r < b_k$, then

$A_r \subseteq \{i \,|\, k+1\le i \le m\}$, so $|A_r| \le m-k$.

On the other hand, $B_r \supseteq \{i \,|\, k \le i \le m\}$, so $|B_r| \ge m-k+1$.

A contradiction. This proves the lemma.

Note: It follows immediately from the lemma that $a_1+ a_2 + \cdots + a_m \geq b_1 +b_2 + \cdots + b_m$.[/quote]

Fuzzylogic proof is very nice as it uses a contradiction argument.

Darij proof is very nice as it is simple to understand.

Thank you very much to both of you to have made clear to me this hard step.

 16 year bump :o

We make some adjustments to start with .
WLOG $\operatorname{max}\{x_1,x_2, \dots x_n\}=\operatorname{max}\{y_1,y_2, \dots y_n\}=1 .$
Next assume WLOG that the $z_i$'s are as small as possible . In other words , assume $z_{i+j}^2 = \operatorname{max}_{i+j=\text{constant}} (x_i \cdot y_j)$ . Hence , $M=1$ .

Now we claim that the following inequality holds true .
Claim :
$$M+ \sum_{k=2}^{2n}z_k \geq \sum_{k=1}^{n}x_k +  \sum_{k=1}^{n}y_k$$.

Proof :

We basically need to show that for all reals $r \in (0,1)$ , the following inequality holds :
$($# terms on LHS $\geq r)$ $\geq$ $($# terms on RHS $\geq r)$

Define sets$$X = \{i : x_i \geq r\}$$$$Y = \{j : y_j \geq r\}$$
Note that
$ z_{i+j}^2 \geq x_iy_j \implies$ $($#terms on LHS $\geq r)$$ \geq |X+Y|+1 .$
(The extra +1 term is due to the extra $M=1$) .

Hence we just need $|X+Y|+1 \geq |X|+|Y|$ , which is a well known result . We are done by AM-GM :)

[b][color=#00f]Remark:[/color][/b] Similar to above proofs, but shows why $|X + Y| + 1 \geq |X| + |Y|$ is well known.

Pick all $z_k$ to be the least possible; that is, for each $k$, let $z_k = \max_{i + j = k} \sqrt{x_iy_j}$. Notice that since this inequality is homogeneous, we may actually scale $\max{(x_1, \ldots , x_n)}$ and $\max{(y_1, \ldots , y_n)}$ both to $1$, so $M = 1$. We wish to prove that\[(M + z_2 + \ldots + z_{2n})^2 \geq 4(x_1 + \ldots + x_n)(y_1 + \ldots y_n)\]but since $(x_1 + \ldots x_n) + (y_1 + \ldots + y_n) \geq 2\sqrt{(x_1 + \ldots + x_n)(y_1 + \ldots y_n)}$ it suffices to show that\[M + z_2 + \ldots + z_{2n} \geq \sum x_i + \sum y_i.\]We will do this by showing that for every real $r \in [0, 1]$, the number of terms in the LHS $\geq r$ is at least the number of terms in the RHS $\geq r$. Let $S$ be the set of indices $s$ for which $x_s \geq r$ and $T$ be the set of indices $t$ for which $y_t \geq r$.

Clearly the RHS has $|S| + |T|$ terms $\geq r$. The LHS has at least $|S + T| + 1$ terms $\geq r$ where $S + T$ is the set of reals of the form $x_i + y_j$ for $i \in S$ and $j \in T$. So it remains to prove that\[|S + T| \geq |S| + |T| - 1\]for all sets $S, T$. We just need to find $|S| + |T| - 1$ distinct elements in $S + T$. Write elements of $S$ and $T$ as $a_1 < \ldots < a_{|S|}$ and $b_1 < \ldots < b_{|T|}$ in increasing order. WLOG $|S| \leq |T|$.

First, consider the sums\[\{x_1 + y_1, y_1 + x_2, x_2 + y_2, \ldots , y_{|S| - 1} + x_{|S|}, x_{|S|} + y_{|S|}\}\]which are clearly all distinct and increasing. There are $2|S| - 1$ of them. Next, we just need to consider\[\{x_{|S|} + y_{|S| +1}, \ldots , x_{|S|} + y_{|T|}\}\]which again are clearly increasing and $x_{|S|} + y_{|S| +1} > x_{|S|} + y_{|S|}$. There are $|T| - |S|$ of them. All of these terms must be distinct because they are arranged in increasing order, hence this is a total of $(2|S| - 1) + (|T| - |S|) = |S| + |T| - 1$ terms, as desired.

Tracking back we may see that our original inequality is true. $\blacksquare$

[i]A different solution:[/i] (it seems a bit long, but in reality it's pretty short to visualize; it seems longer since stuff is written formally not intuitively)
[rule]
Let $x = \max(x_1,\ldots,x_n), y = \max(y_1,\ldots,y_n)$. By scaling, WLOG assume $x = y = 1$. We will more strongly show
$$ M + z_1 + z_2 + \cdots + z_{2n} \ge x_1 + \cdots + x_n + y_1 + \cdots + y_n $$
Consider a bipartite graph $G$ with components $X = \{x_1,\ldots,x_n\}, Y = \{y_1,\ldots,y_n\}$. Place the vertices in that order only, as below:
[asy]
dot("$x_1$",(1,0),dir(-90));
dot("$x_2$",(1.4,0),dir(-90));
label("$\cdots$",(1.6,0),dir(0));
dot("$x_n$",(2,0),dir(-90));
dot("$y_1$",(1,0.5),dir(90));
dot("$y_2$",(1.4,0.5),dir(90));
label("$\cdots$",(1.6,0.5),dir(0));
dot("$y_n$",(2,0.5),dir(90));
[/asy]
[b][color=#f00]Claim: [/color][/b] We can draw $2n-1$ directed edges (segments) in $G$ such that:
[list]
[*] for any vertex $v \ne x$ there is a edge directed towards $v$.
[*] For any edge $e = \overrightarrow{e_1e_2}$, the value of $e_1$ is more than that of $e_2$.
[*] No two segments intersect in their interiors.
[/list]
[b][color=#00f]Proof: [/color][/b] We will draw the edges one by one. Start by drawing $x \to y$. Denote by $S$ the set of vertices which are endpoint of some drawn edge. So initially $S = \{x,y\}$. At any step, choose $v \in G \setminus S$ with max possible value. Note it suffices to draw a edge directed towards $v$ with other endpoint in $S$ and not intersecting any other edges.  FTSOC such a edge cannot be drawn. For segments $s_1,s_2$, write $s_1 \sim s_2$ if $s_1,s_2$ intersect in their interiors, and $s_1 \not\sim s_2$ otherwise. Pick any $a_0$ in different component as of $v$ and consider $a_0v$. There must be some $a_1b_1 \in E(S)$ with $a_0v \sim a_1b_1$. Then we consider $va_1$. In general, if we consider $va_i$ then there is some $a_{i+1}b_{i+1} \in E(S)$ with $va_i \sim a_{i+1}b_{i+1}$. WLOG $a_1$ is to the right of $a_0$. By induction, it is not hard to see for any $i \ge 1$, we have $b_i$ is to the left of $v$ and $a_i$ is to the right of $a_0$ (we have to use no two $a_ib_i$ for $i \ge 1$ intersect).
[asy]
size(200);
dot("$v$",(1,0),dir(-90));
dot("$a_0$",(1,2),dir(90));
dot("$b_1$",(-4,0),dir(-90));
dot("$a_1$",(2,2),dir(90));
draw((1,0)--(1,2),red);
draw((2,2)--(-4,0),green);
dot("$a_2$",(3,2),dir(90));
dot("$b_2$",(-2.5,0),dir(-90));
draw((1,0)--(2,2),red);
draw((3,2)--(-2.5,0),green);
label("$\cdots$",(3.8,2));
label("$\cdots$",(-1.4,0));
dot("$b_i$",(-0.6,0),dir(-90));
dot("$a_i$",(4.6,2),dir(90));
draw((1,0)--(3,2),red);
draw((-0.6,0)--(4.6,2),green);
[/asy]
But this would give us an infinite sequence $\{a_i\}$, contradiction $|S|$ is finite at any point. $\square$

Let $E$ be the edge set creates above. Note for any $x_iy_j \in E$, the values of $i + j$ are all distinct. Thus,
\begin{align*}
& M + z_1 + z_2 + \cdots + z_{2n} \ge 1 + \sum_{\overrightarrow{e_1e_2} \in E} \sqrt{\text{value} (e_1) \text{value} (e_2) } \ge 1 + \sum_{\overrightarrow{e_1e_2} \in E} \text{value} ( e_2) \\
 & x_1 + \cdots + x_n + y_1 + \cdots + y_n = x + \sum_{\overrightarrow{e_1e_2} \in E} \text{value} (e_2)
\end{align*}
This completes the proof of the problem. $\blacksquare$

Another inequality! :D Since the equation is “homogenous”, we can scale max(x_1,…,x_n)=1 and max(y_1,…,y_n)=1, suppose they are x_a and y_b that =1. Because as z_1 increases, the inequality will only get stronger, so it suffices to show the minimum, which is z_k^2=x_iy_(k-i) s.t. the product of z_k^2 is the maximum of all possible products. Then we have M=1. Now, we can show that $M+z_1+…+z_{2n}\ge x_1+…+x_n+y_1+…+y_n$ by showing that for every value w, the LHS has at least as many values greater or equal to w than the RHS does. This is easy since if any arbitrary x_c and y_d are each greater than r ($0\le r\le 1$), then it also makes values z_(a+d) and z_(b+c) greater than r, since they are multiplied by 1. There there is a 1-1 correspondence, two values of z for every 1 value of each x and y. From here, we would have $(LHS/2n)^2\ge (RHS/2n)^2=(((x_1+…+x_n)/n+(y_1+…+y_n)/n)/2)^2\ge (sqrt{(x_1+…+x_n)/n(y_1+…+y_n)/n})^2=(x_1+…+x_n)/n(y_1+…+y_n)/n$, as desired. $\blacksquare$


function
linear algebra
matrix
algebra
system of equations
algebra proposed

Find all the functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that :

$ x^2f(x)\plus{}f(2\minus{}x)\equal{}2x^3\plus{}x^2\minus{}2x\plus{}5\ ,\ (\forall)x\in \mathbb{R}$

[quote="alex2008"]Find all the functions $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that :

$ x^2f(x) \plus{} f(2 \minus{} x) \equal{} 2x^3 \plus{} x^2 \minus{} 2x \plus{} 5\ ,\ (\forall)x\in \mathbb{R}$[/quote]
Substitute $ x \to 2\minus{}x$:
$ (2\minus{}x)^2f(2\minus{}x) \plus{} f(x) \equal{} 2(2\minus{}x)^3 \plus{} (2\minus{}x)^2 \minus{} 2(2\minus{}x) \plus{} 5$
Hence we can easily find $ f(x)$ or $ f(2\minus{}x)$ for all $ x$...

This is somewhat ugly but here it is anyway.
[hide]Plugging in $ x$ yields:
$ x^2f(x) \plus{} f(2 \minus{} x) \equal{} 2x^3 \plus{} x^2 \minus{} 2x \plus{} 5$

Plugging in $ 2 \minus{} x$ yields:
$ (2 \minus{} x)^2f(2 \minus{} x) \plus{} f(x) \equal{} 2(2 \minus{} x)^3 \plus{} (2 \minus{} x)^2 \minus{} 2(2 \minus{} x) \plus{} 5$ $ \equal{} \minus{} 2x^3 \plus{} 13x^2 \minus{} 26x \plus{} 21$

This yields a system of equations for $ f(x)$ and $ f(2 \minus{} x)$:

$ \begin{bmatrix} x^2 & 1 \\
1 & (2 \minus{} x)^2 \end{bmatrix} \begin{bmatrix} f(x) \\
f(2 \minus{} x) \end{bmatrix}$ $ \equal{} \begin{bmatrix} 2x^3 \plus{} x^2 \minus{} 2x \plus{} 5 \\
\minus{} 2x^3 \plus{} 13x^2 \minus{} 26x \plus{} 21 \end{bmatrix}$

Solving this system yields $ f(x) \equal{} 2x \plus{} 1$
for all $ x$ such that $ (x)^2(2 \minus{} x)^2 \minus{} (1)(1) \neq 0 \Rightarrow [x(x \minus{} 2)]^2 \neq 1$ $ \Rightarrow x^2 \minus{} 2x \pm 1 \neq 0 \Rightarrow x \not\in \{1 \minus{} \sqrt {2}, 1, 1 \plus{} \sqrt {2} \}$.

Plugging in $ x \equal{} 1$ into the original equation gives $ 2f(1) \equal{} 6 \Rightarrow f(1) \equal{} 3 \equal{} 2(1) \plus{} 1$.

Plugging in $ x \equal{} 1 \plus{} \sqrt {2}$ yields $ (3 \plus{} 2\sqrt {2})f(1 \plus{} \sqrt {2}) \plus{} f(1 \minus{} \sqrt {2}) \equal{} 20 \plus{} 10\sqrt {2}$ (A).

Plugging in $ x \equal{} 1 \minus{} \sqrt {2}$ yields $ (3 \minus{} 2\sqrt {2})f(1 \minus{} \sqrt {2}) \plus{} f(1 \plus{} \sqrt {2}) \equal{} 20 \minus{} 10\sqrt {2}$ (B).
However, multiplying (B) by $ 3 \plus{} 2\sqrt {2}$ yields (A).

So, if we arbitrarily define $ f(1 \plus{} \sqrt {2}) \equal{} k \in \mathbb{R}$, then $ f(1 \minus{} \sqrt {2}) \equal{} 20 \plus{} 10\sqrt {2} \minus{} (3 \plus{} 2\sqrt {2})k$

Therefore:
$ f(x) \equal{} 2x \plus{} 1$ for all $ x \neq 1\pm\sqrt {2}$, $ f(1 \plus{} \sqrt {2}) \equal{} k$, and $ f(1 \minus{} \sqrt {2}) \equal{} 20 \plus{} 10\sqrt {2} \minus{} (3 \plus{} 2\sqrt {2})k$[/hide]


inequalities
calculus
integration
function
real analysis
real analysis unsolved

let f, g be 2 continuous functions on [a,b],
$ f(x) \neq 0$ for all $ x \in[a,b]$ and m,M satisfy: $ m\leq \frac{g(x)}{f(x)} \leq M$ for all $ x \in [a,b]$
prove that: $ \int_a^bg^2(x)dx \plus{}Mm\int_a^bf^2(x)dx \leq (M\plus{}m)\int_a^bf(x)g(x)dx$

$ g^2\minus{}(M\plus{}m)fg\plus{}Mmf^2\equal{}(g\minus{}Mf)(g\minus{}mf)\le 0$ pointwise.

We have  $ f^2(x)(\frac {g(x)}{f(x)} \minus{} m)(\frac {g(x)}{f(x)} \minus{} M) \leq 0 , \forall x \in [a,b]$ or $ (g(x) \minus{} mf(x))(g(x) \minus{} Mf(x)) \leq 0 ,\forall x \in [a,b]$
 $ \int_{a}^{b}{(g(x) \minus{} mf(x))(g(x) \minus{} Mf(x))dx} \equal{} \int_{a}^{b}g^{2}(x)dx \plus{} Mm\int_{a}^{b}f^{2}(x)dx \minus{} (M \plus{} m)\int_{a}^{b}f(x)g(x)dx \leq o$
From here $ \int_{a}^{b}g^{2}(x)dx \plus{} Mm\int_{a}^{b}f^{2}(x)dx\leq (M \plus{} m)\int_{a}^{b}f(x)g(x)dx$


solve the following system:

$ \sum{x_{i}}\equal{}1$

$ \sum{x_{i}^{\alpha}}\equal{}1$  $ (\alpha\in{N})$

good luck! :)

Sorry :blush: ^^:

$ \sum_{i \equal{} 1}^{n}{x_{i}} \equal{} 1$

$ \sum_{i \equal{} 1}^{n}{x^{\alpha}_{i}} \equal{} 1$

$ \alpha$ is a constant($ \in$IN*-{1})

I think it's clear now  :wink:

hello NoW that's OK !!!!  :)

than it's obvious that's $ S\equal{}\{for\: i \in \{1;2;...;n\} \: it\:  found \: one \: j \in \{1;...;n\}: (x_j)\equal{}1 \: and \: x_{i \neq j}\equal{}0 \}$ are a solutions of your system.

let now $ x_i \equal{} (\minus{}1)^{i\minus{}1}$ for  $ impair$ numbers $ n$ and $ \alpha$ than $ S\equal{}\{x_i\equal{}(\minus{}1)^{i\minus{}1} \ with \ (n;\alpha) \in (2\mathbb{N}\plus{}1)^2 \}$ are also a solutions of your system....

to be continued.....
______________________________________________________________
L.A.H

we consider that we have just $ (i \leq 2)$:

$ x_1 + x_2 =1$

$ x_1^\alpha + x_2^\alpha =1$ $ (S)$

than

$ (S)$ $ \Longleftrightarrow$

$ x_2 = (\sum_{k=0}^{\alpha -1} x_1^k)^{\frac{1}{\alpha -1}}$ with $ x_1 \geq 0$ and $ \alpha \succ 1$

that we have infinity of solutions $ \{(x_1 ;  (\sum_{k=0}^{\alpha -1} x_1^k)^{\frac{1}{\alpha -1}})\}$

the same if $ i \geq 2$

than conclusion:

 your system has infinity of solutions ..
__________________________________________________________
L.A.H


On each vertex of a regular $ n\minus{}$gon there was a crow. Call this as initial configuration. At a signal, they all flew by and after a while, those $ n$ crows came back to the $ n\minus{}$gon, one crow for each vertex. Call this as final configuration. Determine all $ n$ such that: there are always three crows such that the triangle they formed in the initial configuration and the triangle they formed in the final configuration are both right-angled triangle.

the condition is satisfied iff $ n$ is even. the "only if" part is obvious.

the "if" part: if $ n$ is even, then let $ A, B$ be two diametrically opposed crows in the initial configuration. if $ B$ is also diametrically opposite $ A$ in the final configuration, then clearly $ A$, $ B$ and any third crow will satisfy the condition. if $ C\neq B$ is the crow opposite $ A$ in the final configuration, then $ A, B, C$ form a triangle in the initial configuration as well as in the final configuration.


[quote="GameBot"]The price of a radio was discounted $ 20\%$ on the first day of
a sale. The sale price was then discounted an additional $ 20\%$
on the second day. What percent is the combined discount?[/quote]

0.8*0.8=0.64.  100-64=36 percenttttttt

Let the original price be 100
After 1 day, its 100*.8=80
After 2 days, its 80*.8=64
64 out of 100 dollars is 36 percent off.


[quote="Zootieroaz"]Let a, b, c and d be integers with a<2b, b<3c, c<4d. If d<100, what is the largest value for a ?[/quote]

[hide]
$c = 4d - 1$
$b = 3c - 1$
$a = 2b - 1$
$a = 2(3(4d - 1) - 1) - 1 = 24d - 9 =2367$
[/hide]
 :blush:

but inside the solutions there are

a, 2367
b, 2375
c, 2391
d, 2399
d, 2400

sry but now i see that u mad a mistake cos that will be 24d-9 and that is 2367 but thx for the help!


vector
topology
probability
analytic geometry
linear algebra
matrix
group theory

In Smooth Manifolds, Lee states in example 1.24 that in a vector space $ V$ of dimension $ n$, given two $ k$-dim vector spaces $ P$ and $ P'$, it is possible to find a vector space $ Q$ of dimension $ n\minus{}k$ such that $ P \cap Q \equal{} P' \cap Q \equal{} 0$.

I've been trying to verify this statement, but I can't seem to come up with a proof.  I figure it's just a dimension thing, but nothing seems to be working.  Can anyone offer me a hint?

Two arguments:

If we choose a space of dimension $ n\minus{}k$ at random (assume the field is $ \mathbb{R}$ or $ \mathbb{C}$), it intersects $ P$ trivially with probability $ 1$ and intersects $ P'$ trivially with probability $ 1$.

Choose $ n\minus{}k$ vectors $ v_i$ that span a space complementary to $ P$. For each $ i$, there is some $ w_i\in P$ such that $ v_i\plus{}w_i$ is not in $ P'$. Those $ v_i\plus{}w_i$ span the $ Q$ we seek.

I understand the almost everywhere argument, but I'd rather try to avoid it and do something a bit more constructive.

For the 2nd argument, this is what I've been trying to do, but I fail to see how $ P' \cap Q \equal{} 0$.  Just because each of the basis vectors is not in $ P'$ does not mean their span intersects $ P'$ trivially.  I see that I somehow need to use the fact that $ V \equal{} P \oplus P^c$, but not quite sure where this comes in.

Put it in convenient coordinates: let the $ v_i$ be the first $ n\minus{}k$ basis vectors, and let $ Q$ be the span of the last $ k$ basis vectors. Then $ v_1\plus{}w_1$ looks like $ \begin{bmatrix}1&0& \cdots& 0& ?& \cdots& ?\end{bmatrix}^T$ and so on; put the $ v_i\plus{}w_i$ in a matrix for $ \begin{bmatrix}I\\?\end{bmatrix}$ in block form. Now, how could you possibly get a linear combination of the columns that zeros out the first $ n\minus{}k$ rows? It has to be the zero combination, and thus the intersection is trivial.

Forgive me if I'm being dense, but an element of $ P'$ need not have the first $ n\minus{}k$ rows zero.  $ \{v_i\}$ forms a basis of $ P^c$ which is not a complement to $ P'$.

Of course. That's why we add the $ w_i$; $ P'$ can't cover all of $ v_i\plus{}P$, since the elements of $ v_i\plus{}P$ span a space of dimension $ k\plus{}1$.

More generally, if $ \dim(P) \equal{} k$, $ \dim(P') \equal{} k'$, find $ Q$ with $ \dim(Q) \equal{} n \minus{} \max(k,k')$ (this is trivially equivalent to actual problem, since we may complete the smaller to the dimension of the larger).

But over $ \mathbb{R}$ or $ \mathbb{C}$ the result remains true for (countable) infinitely many $ k$-subspaces $ P$... However, over $ \mathbb{F}_2$ we cannot improve over two, for example $ V \equal{} \mathbb{F}_2^2$, $ P\equal{}\{(0,0), (1,0)\}$, $ P'\equal{}\{(0,0),(1,0)\}$; then the only $ Q$ is $ Q\equal{}\{(0,0),(1,1)\}$, and we cannot take even a third $ P''$ (for example $ P'' \equal{} Q$, since then we have no vectors left!).

jmerry, in post #4, you asked:

[quote]Now, how could you possibly get a linear combination of the columns that zeros out the first $ n \minus{} k$ rows?[/quote]

Concluding that only coefficients all being zero will suffice.  However, this does not show that $ P' \cap Q \equal{} 0$ because it need not be that the entries in the first $ n \minus{} k$ rows be all zero for an element in $ P'$.

Post #4 only proved that $ \{v_i \plus{} w_i\}$ are linearly independent.

Here is a counter example to your construction, jmerry.  Hopefully I didn't screw up something simple...

$ n \equal{} 4$, $ k \equal{} 2$, standard Euclidean basis

$ P \equal{} \langle (1, 1, 0, 0), (1, 0, 1, 0) \rangle$
$ P' \equal{} \langle (0, 0, 1, 1), (0, 1, 0, 1) \rangle$

The complement of $ P$ is give by:

$ P^c \equal{} \langle (0, 0, 0, 1), (0, 1, 1, 0) \rangle$

So from your construction

$ v_1 \equal{} (0, 0, 0, 1)$ and $ v_2 \equal{} (0, 1, 1, 0)$

Now we may take $ w_1$ and $ w_2$ to be both zero since $ v_1$ and $ v_2$ are not in $ P'$.  This gives

$ Q \equal{} \langle (0, 0, 0, 1), (0, 1, 1, 0) \rangle$

But now the vector $ (0, 1, 1, 2)$ is in $ Q$, but it's also in $ P'$.

Okay, that idea is broken and I don't see how to fix it.

By the way, this belongs in the linear algebra forum, even if it did come from a manifold application.

Probably the easiest proof to understand is the following. Let us use the classical

[color=blue][b]Lemma[/b]. No group can be the union of two of its proper subgroups.[/color]

[i]Proof[/i]. Let $ G$ be a group (we will use additive notation), and $ A$ and $ B$ be two of its proper subgroups. Assume $ G \equal{} A \cup B$. Also assume for the moment that $ a \in A \setminus B$ and $ b \in B \setminus A$. Then $ a \plus{} b \in G$, but $ a \plus{} b \in A$ implies $ b \in A$ ($ a \plus{} b \equal{} a' \in A$, so $ b \equal{} ( \minus{} a) \plus{} a' \in A$), while $ a \plus{} b \in B$ implies $ a \in B$, both contradictions. Therefore either $ B \subseteq A$, when $ G \equal{} A$, or $ A \subseteq B$, when $ G \equal{} B$, against the given that the subgroups are proper. $ \quad \blacksquare$

Now, our $ k$-dimensional subspaces $ P$ and $ P'$ are also proper additive subgroups of the $ n$-dimensional vector space $ V/K$. According with the Lemma, $ P \cup P' \subsetneq V$, hence there exists $ v_1 \in V \setminus (P \cup P')$. Take $ P_1 \equal{} P \oplus Kv_1$ and $ P'_1 \equal{} P' \oplus Kv_1$, both new subspaces of dimension $ k \plus{} 1$. Repeat this operation until finding vectors $ v_1, v_2, \ldots, v_{n \minus{} k}$ (or reason by induction after $ n \minus{} k$). Clearly these vectors are linearly independent, and the $ (n \minus{} k)$-dimensional subspace $ \displaystyle Q \equal{} \bigoplus_{i \equal{} 1}^{n \minus{} k} Kv_i$ spanned  by them intersects both $ P$ and $ P'$ in $ 0$ only.

Slick proof, and using group theory too!  I love it.  Thanks.

A solution using basis:

Let $ b_1,...,b_k$ be a basis for $ P$. Re-order this list into the basis $ p_1,...,p_m,...,p_k$ such that the first $ m$ elements are not in $ P'$, where $ m\geq 0$. Then, extend the basis $ p_1,...,p_k$ to basis $ p_1,...,p_k,...,p_{k\plus{}m}$ of $ P\cup P'$ (I'll clarify this process if I'm requested to). Then, let $ e_1,...,e_{n\minus{}k\minus{}m}$ be a basis of $ V\minus{}(P\cup P')$. Then, note that $ \{p_1\plus{}p_{k\plus{}1},p_2\plus{}p_{k\plus{}2},...,p_{m}\plus{}p_{k\plus{}m}, e_1,...,e_{n\minus{}k\minus{}m}\}$ is a linearly independent set with $ n\minus{}k$ elements. Call its span $ Q$. It is now clear that $ P\cap Q \equal{} P'\cap Q \equal{} \emptyset$.


floor function
integration
trigonometry
combinatorics solved
combinatorics

Given that both $a_i,b_i\in \{1,2,3,4,5,6,7,8,9,0\}$
Find the number of different solutions depending on $n$ of the following equaltion:
$a_1+a_2+a_3+...+a_n=b_1+b_2+b_3+...+b_n=k$ where $k$ takes all values $0\le k\le 9n$
[i]For example for $n = 1$ for each k ($0\le k\le 9$) equation has just one solution sothe answer is 10 and the solutions are $(a_1,b_1)=(1,1);(2,2);(3,3);...;(0,0)$[/i]


ratio
trigonometry
Putnam
inequalities unsolved
inequalities

Five distinct points given on the plane.Let  d1 denotes maximum distance between any five points  and  d2 denotes minumum  distance any five points.Find the ratio of  d1/d2.

That ratio depends on how the 5 points are arranged (ie it's not always the same). Maybe you want the minimum possible ratio (the supremum is $ \plus{}\infty$). I think it's $ 2\cos \frac \pi 5$.

In general, is it true that the minimum ratio between the shortest and the longest side is no smaller than

$ 2 \cos {\frac {\pi}{n}}$

where $ n$ is the number of distinct points?

(EDIT: This is a generalization of 1964 Putnam A1

and actually, this is a weak bound, as in the case with $ n \equal{} 6$ the minimum ratio is $ 2 \sin \frac{2 \pi}{5} > \sqrt{3}$, in the case of a regular pentagon with a central point.)


Maybe it's so easy that no one has replied.
We have $ \sum_{j=1}^{k}log(\frac{k^{2}+j}{k^{2}-j})=\sum_{j=1}^{k}log(1+\frac{2j}{k^{2}-j})$ and $ \frac{2j}{k^{2}-j}-\frac{1}{2}(\frac{2j}{k^{2}-j})^{2}<log(1+\frac{2j}{k^{2}-j})<\frac{2j}{k^{2}-j}$.
Because $ \frac{2j}{k^{2}-j}\leq \frac{k}{k^{2}-k}=\frac{1}{k-1}$,$ \sum_{j=1}^{k}(\frac{j}{k^{2}-j})^{2}\leq \frac{k}{(k-1)^{2}}$.This means when $ k\to \infty$, the sum of this part $ \frac{1}{2}(\frac{2j}{k^{2}-j})^{2}$ tends to $ 0$.
As to the sum of $ \frac{2j}{k^{2}-j}$,we have $ \frac{2}{k^{2}-1}\sum_{j=1}^{k}j<\sum_{j=1}^{k}\frac{2j}{k^{2}-j}<\frac{2}{k^{2}-k}\sum_{j=1}^{k}j$,from which it is readily deduced that when $ k\to \infty$,$ \sum_{j=1}^{k}\frac{2j}{k^{2}-j}$ tends to $ 1$.
The  desired result obtained.


calculus
integration
trigonometry
calculus computations

[quote="nguyenvuthanhha"][u]  [i] Problem  [/i] [/u]     [i]Evaluate :[/i]

        $ A \ \equal{} \ \int_{0}^{1} \frac { 4x \ \plus{} \ 3 }{ ( x^2 \minus{} x \plus{} 1 )^{2} } dx$[/quote]

$ A\equal{}2 \int_{0}^{1} \frac {  ( x^2 \minus{} x \plus{} 1 )'  }{ ( x^2 \minus{} x \plus{} 1 )^{2} }  \;\plus{}\;5
 \left(\frac{2}{\sqrt 3}  \right)^3\int_{0}^{1} \frac{  \left[\frac{2}{\sqrt 3} \left(x\minus{}\frac{1}{2}\right) \right]' } { \left( 1\plus{} \left[\frac{2}{\sqrt 3}  \left(x\minus{}\frac{1}{2}\right)\right]^2    \right)^{2} }$


function
inequalities
Stanford
college
real analysis
real analysis unsolved

Let $ S$ be a linear subspace of $ C([0,1])$ such that
\[ \|f\|_{L^\infty([0,1])} \le \lambda \|f\|_{L^2([0,1])}
\]
for any $ f \in S$. Let $ v_1,\ldots,v_n \in S$ be orthonormal functions in $ L^2([0,1])$. Prove that

(a) $ \sum_{k\equal{}1}^n |v_k(x)|^2 \le \lambda^2$ for any $ x\in[0,1]$;
(b) $ \dim S \le \lambda^2$.

This question seems to have problem. One can pick $ v_k$ to be very large at one point so that the inequality in (a) is wrong.

Probably you forgot about the fact that $ v_k$ are in $ S$ that has the property that
\[ \|f\|_{L^\infty} \le \lambda \|f\|_{L^2}
\]
for any $ f \in S$. The problem has to be correct since it has been offered on Qualifying Exams at Stanford Univ., and I have a solution for this problem.

Sorry for overlooking the continuity condition last time.

(a) For constants $ t_1,...,t_n$, apply the given inequality to $ f \equal{} t_1v_1 \plus{} ... \plus{} t_nv_n$. By continuity, we have $ sup |t_1v_1 \plus{} ... \plus{} t_nv_n| \leq \lambda \sqrt {|t_1|^2 \plus{} ... \plus{} |t_n|^2}$
In particular, for any $ x$, $ |t_1v_1(x) \plus{} ... \plus{} t_nv_n(x)| \leq \lambda \sqrt {|t_1|^2 \plus{} ... \plus{} |t_n|^2}$
Now substitute $ t_i \equal{} \bar{v_i(x)}$

(b) Integrate both sides in (a) on [0,1], we see that $ k \leq \lambda$ for any orthonormal set of $ k$ elements. Then the result is obvious.

Yes, my solution is the same. And, despite the fact that the problem is not hard, I couldn't solve it for a long time.  :blush:


The number $\sqrt{7 + 4 \sqrt 3} + \sqrt{7 - 4 \sqrt 3}$ equals what rational number?

[hide] first, i had $a=7$ and $b=4\sqrt{3}$

we then have the number

$\sqrt{a+b}+\sqrt{a-b}=x$

square both sides

$a+b+2\sqrt{a^2-b^2}+a-b=x^2$

Then we have $2(a+\sqrt{a^2-b^2})=x^2$

Now substitute $a=7$ and $b=4\sqrt{3}$

and we get $2(7+1)=x^2$

then $x=4,-4$
However, by squaring in the beginning, we created an extra answer. Looking back at the problem, it is obvious that the number must be positive. Therefore, the answer is 4.[/hide]

No need for so many steps.
 $(\sqrt{7 + 4 \sqrt 3} + \sqrt{7 - 4 \sqrt 3})^2 = 7+4\sqrt3 + 7-4\sqrt3 + 2(7^2 - (4\sqrt3)^2) = 16$,
which implies the given number $ = 4$

$sqrt(3+4+4sqrt3)$ and same the other with the other one so
=$sqrt(sqrt3+2)$ so the answer will come as $4$
easiest way :rotfl:


Hi,Do you mean $ \cos(\cos x) \geq \sin x ,$   $ \forall x \in [0,\frac {\pi}{2}]$  ?


I've seen that Belgian archaeologists have taken pottery from ancient Pompeii and played the grooves back like a record to get the sounds of the pottery workshop, including laughter!?

via the website of a film-maker:
http://www.bilgesehir.com/c_filmo_principale/e_vases_sonores/Bilge%20Sehir%20-%20Vases%20sonores.htm

Click "Pour visionner le video" to see a video which contains a sample of the audio.......

Wow, that's amazing. What's even more amazing is that I was able to more or less follow what they were saying, even though I'm in French 2. I guess I'm wrong about having to take years upon years to be able to communicate (semi)normally.

[quote="Valentin Vornicu"]You do realize that ancient people in Pompeii spoke Latin, as opposed to French, and you probably would not understand 95% of what they were saying? :)[/quote]

They being the commentators.

[quote="Setofallsets"]The whole deal is a hoax anyway (an old April fool's joke, [url=http://www.museumofhoaxes.com/hoax/weblog/permalink/ancient_pottery_recorded_audio/]apparently[/url]).[/quote]

Bah, that's terrible. I believed it.


I am taking the AMC12 on Tuesday. I bought AoPS 1+2 about a month ago and I planned on studying from them, but I was really busy and did not have time :oops: .  I have been taking old AMC tests this weekend to prepare.  What else can I study from today and tomorrow that will help me ?  What are the the most important chapters in AoPS that I can learn before the test?

Thanks

There's no "most important" chapter.  Read whatever sections you feel you need to work on.  Getting a lot of number theory questions wrong?  Read the chapter on number theory.  Really good at geometry?  Don't read that chapter.

Read my guide, and one part mentions this.

http://www.hiddendiscoveries.com


A tube containing a solution is closed at the bottom by a semipermeable membrane. It is then placed in a cup of pure water. Which solution/temperature would make the solution in the tube rise the fastest?

a) 10% glucose, 37 Celcius
b) 6% salt, 37 Celcius
c) 6% glucose, 60 Celcius
d) 14% salt, 60 Celcius

I believe its D; the osmotic pressure is greatest.


prove that if x,y,z>=0, show that 3xyz+ summation{x^3}>=summation {x^2y+y^2x}

i wonder why it is equivalent to proving summation{x(x-y)(x-z)}>=0

It is equivalent because:
$ LHS\minus{}RHS\geq 0$
$ \sum_{cyc}x^{3}\plus{}xyz\minus{}x^{2}y\minus{}x^{2}z\geq 0$
$ \sum_{cyc}x(x\minus{}y)(x\minus{}z)\geq 0$ which is true by Schur's inequality :wink:

Schur Inequlity:
 \[ \sum_{cyc} x^r(x\minus{}y)(x\minus{}z) \geq 0\] for any nonnegative $ x, y, z, r$  :roll:

[quote="chess"]Schur Inequlity:
\[ \sum_{cyc} x^r(x \minus{} y)(x \minus{} z) \geq 0
\]
for any nonnegative $ x, y, z, r$  :roll:[/quote]

Schur's inequality $ a,b,c$ nonnegative numbers and $ r \in R$ then
\[ \sum_{cyc} a^r(a\minus{} b)(a \minus{} c) \geq 0\]
inequality holds.

[quote="morheli"]
i wonder why it is equivalent to proving summation{x(x-y)(x-z)}>=0[/quote]

That's what you get when you move the RHS to the LHS and factor.


calculus
integration
calculus computations

After evaluating the integral (by parts),  we have $I=\frac{a^{4}+4}{a^{2}}\Rightarrow I' = \frac{2(a^{4}-4)}{a^{3}}=0\Rightarrow a =\sqrt{2}$  And thus $I_{min}= 4$.

[quote="kunny"]That's O.K. You can use A.M.-G.M. :)[/quote]

I always forget to use AM-GM  :) . By AM-GM, we have $\frac{a^{4}+4}{2}\ge \sqrt{4a^{2}}\Rightarrow a^{4}+4\ge4a^{2}\Rightarrow\frac{a^{4}+4}{a^{2}}\ge 4$.


geometry
trigonometry
angle bisector
trig identities
Law of Sines
Law of Cosines

Let ABC is a triangle satisfying that A + C=60, BC=a,AB=c,AD is a bisector of angle A.find AD by a and c :lol:

[hide]
expressing in terms of area:
$[ABC]=\frac{1}{2}ac\sin{120}$
$[ABD]=\frac{1}{2}cd\sin{60}$
$[BDC]=\frac{1}{2}ad\sin{60}$

$[ABC]=[ABD]+[BDC]$
$\frac{1}{2}ac\sin{120}=\frac{1}{2}cd\sin{60}+\frac{1}{2}ad\sin{60}$
$d=\frac{ac}{a+c}$
[/hide]
EDIT:  :blush: oh. haha i guess not THAT easy :(

[quote="TRAN THAI HUNG"]Let $\triangle ABC$ satisfy $\angle A + \angle C=60, BC=a,AB=c.$ $AD$ bisects $\angle A$. Find $AD$ in terms of $a$ and $c$ :lol:[/quote]
[hide="Well..."]
[quote="AoPS Vol 2 page 39"]
To find the length of an angle bisector, we recall the Angle Bisector Theorem, which gives n/m=b/c. We can use thsi relation and Stewart's Theorem (and a bit of algebra) to find
                   $d^2=ac(1-\frac{b^2}{(a+c)^2})$
Solve Stewart's Theorem for d and use the Angle Bisector Theorem in the form m/(m+n)=c/(a+c)
[/quote]
So the problem tells us $\angle A+\angle C=60 \implies \angle B=120$. $BC^2=a^2+c^2-2(a)(c)\cos{\angle A}=a^2+c^2+(a)(c) \implies BC=\sqrt{a^2+ac+c^2}$. Plugging this into $d^2=ac(1-\frac{b^2}{(a+c)^2})$ gives ${d^2=ac(1-\frac{a^2+ac+c^2}{(a+c)^2}) \implies \boxed{d=AD=\sqrt{ac(1-\frac{a^2+ac+c^2}{(a+c)^2})}}=\frac{ac}{a+c}}$
[/hide]

ok...
[hide]
using law of sines, angle bisector theorem, and stewarts we can find that $AD^2 = bc*\left[1-\left(\frac{a}{b+c}\right)^2\right]$, since you only want it in terms of $a$ and $c$, and we know that $\angle B = 120$, we have $b^2 = a^2 + c^2 + a*c$ by law of cosines, then we just plug this in the formula
[/hide]

[quote="me@home"][quote="TRAN THAI HUNG"]Let $\triangle ABC$ satisfy $\angle A + \angle C=60, BC=a,AB=c.$ $AD$ bisects $\angle A$. Find $AD$ in terms of $a$ and $c$ :lol:[/quote]
[hide="Well..."]
[quote="AoPS Vol 2 page 39"]
To find the length of an angle bisector, we recall the Angle Bisector Theorem, which gives n/m=b/c. We can use thsi relation and Stewart's Theorem (and a bit of algebra) to find
                   $d^2=bc(1-\frac{a^2}{(b+c)^2})$
Solve Stewart's Theorem for d and use the Angle Bisector Theorem in the form m/(m+n)=c/(b+c)
[/quote]
So the problem tells us $\angle A+\angle C=60 \implies \angle B=120$. $BC^2=b^2+c^2-2(b)(c)\cos{\angle A}=b^2+c^2+2(b)(c)=(b+c)^2 \implies BC=b+c$. Plugging this into $d^2=bc(1-\frac{a^2}{(b+c)^2})$ gives $d^2=bc(1-\frac{(b+c)^2}{(b+c)^2})=bc(1-1)=0 \implies \boxed{d=AD=0}$
So we have a degenerate triangle (we knew that after find $b+c=BC$) so the bisector is zero... easy problem if I saw the trick first before looking up the formula... :roll:[/hide][/quote]

I think you got this wrong....you should find $b$ in terms of $a$ and $c$ instead of finding $a$.  Because you don't know $b$.  Also, you did something wrong in your formula, because $BC^2=b^2+c^2-2(b)(c)\cos{\angle A}=b^2+c^2+2(b)(c)$ if and only if $\angle A = 180$, but A must be smaller than 60 becaseu $A + C = 60$, contradiction.

[quote="me@home"][quote="TRAN THAI HUNG"]Let $\triangle ABC$ satisfy $\angle A + \angle C=60, BC=a,AB=c.$ $AD$ bisects $\angle A$. Find $AD$ in terms of $a$ and $c$ :lol:[/quote]
[hide="Well..."]
[quote="AoPS Vol 2 page 39"]
To find the length of an angle bisector, we recall the Angle Bisector Theorem, which gives n/m=b/c. We can use thsi relation and Stewart's Theorem (and a bit of algebra) to find
                   $d^2=bc(1-\frac{a^2}{(b+c)^2})$
Solve Stewart's Theorem for d and use the Angle Bisector Theorem in the form m/(m+n)=c/(b+c)
[/quote]
So the problem tells us $\angle A+\angle C=60 \implies \angle B=120$. $BC^2=b^2+c^2-2(b)(c)\cos{\angle A}=b^2+c^2+2(b)(c)=(b+c)^2 \implies BC=b+c$. Plugging this into $d^2=bc(1-\frac{a^2}{(b+c)^2})$ gives $d^2=bc(1-\frac{(b+c)^2}{(b+c)^2})=bc(1-1)=0 \implies \boxed{d=AD=0}$
So we have a degenerate triangle (we knew that after find $b+c=BC$) so the bisector is zero... easy problem if I saw the trick first before looking up the formula... :roll:[/hide][/quote]Try to find the mistake in your solution. Hint: check carefully the [b]angles[/b] with a picture.

I got the same answer as pkerichang's but let's see what TRAN THAI HUNG has to say ...

I must admit, the formula seems nasty and I can't find a way to simplify it... :?

angle bisector theorem: $BD=\frac{ac}{b+c}$

law of cosines: $b^2=a^2+c^2+ac$

again, $AD^2=c^2+BD^2+c*(BD)$

plug in and you are done

[quote="me@home"]I know of at least one mistake I made-- cos120=-1/2, not -1... so it would be the same as what everyone is saying... a^2+ab+b^2 not (a+b)^2[/quote]That is not the mistake, the mistake in your solution is ... you don't know what angle A is because we only know A+C=60.

[quote="me@home"][hide]
So the problem tells us $\angle A+\angle C=60 \implies \angle B=120$. $BC^2=a^2+c^2-2(a)(c)\cos{\angle A}=a^2+c^2+(a)(c) \implies BC=\sqrt{a^2+ac+c^2}$. Plugging this into $d^2=ac(1-\frac{b^2}{(a+c)^2})$ gives ${d^2=ac(1-\frac{a^2+ac+c^2}{(a+c)^2}) \implies \boxed{d=AD=\sqrt{ac(1-\frac{a^2+ac+c^2}{(a+c)^2})}}=\frac{ac}{a+c}}$
[/hide][/quote]

First, $BC=a$, as was stated in the first post. Next, the $d$ you found is the angle bisector of $B$. Draw and label a diagram, and apply formulas correctly.

I also ended up plugging the law of cosines expression for $b$ into the angle bisector formula and gave up trying to simplify it.[/hide]


prove that if n divides( 2^(n))+1 then n=3 or 9 divides n

Let $n>1$, a and let $p$ be least prime divisor of $n$. It's ok that $p>2$. Now we have $p \mid 2^{2n}-1$ and $p \mid 2^{p-1}-1$, so $p \mid 2^{(2n,p-1)}-1$. From this we have $p \mid 2^{2}-1$, hence $p=3$.

Then we have $n=3$ or $n=3d$ where $d>1$. Now take $n \neq 3$, and let $q$ be least prime divisor of $d$.

As above we have $q \mid 2^{2n}-1$ so $q \mid 2^{6d}-1$ and $q \mid 2^{q-1}-1$ hence $q \mid 2^{(6d,q-1)}$ and we have following casses.

1) $(6d,q-1)=2$

	There will be $q \mid 2^{2}-1$ hence $q=3$ and we have $9 \mid n$.

2) $(6d,q-1)=6$

	There will be $q \mid 2^{6}-1$ , hence $q \mid 63$ and there we will have $q=3$ or $q=7$. If $q=3$ it will be $9 \mid n$, and we proved that case. If $q=7$ then it is suposed to be $7 \mid 2^{3d}+1$ or $7 \mid 8^{d}+1$ which leads to contradiction beacuse $8^{d}\equiv_{7}1$.

qed.


To cross a shallow river of width 14 meters without getting my feet wet, I would like stepping stones placed in a river so that they form a straight path perpendicular to the riverbank. Stones can be placed 1, 2, 3, ... 12, 13 feet away from the shore. Assuming I can only make a leap of at most 4 meters, in how many ways can I place between 4 and 12 stones in these positions.

[hide="Answer"]
It should be 2871. Or am I incorrect?
[/hide]

[quote="Caelestor"]Stones can be placed 1, 2, 3, ... 12, 13 feet away from the shore.
[/quote]

This is supposed to be meters, not feet, right?

If it is, then this is a recursion. The number of ways to place any number of stones in positions 1, 2, ...n-1 for a n-meter wide river (Let this be denoted by $ a_n$) is $ a_{n\minus{}1}\plus{}a_{n\minus{}2}\plus{}a_{n\minus{}3}\plus{}a_{n\minus{}4}$. $ a_1\equal{}1$, $ a_2\equal{}2$, $ a_3\equal{}4$, and $ a_4\equal{}8$. Therefore:
$ a_5\equal{}1\plus{}2\plus{}4\plus{}8\equal{}15$
$ a_6\equal{}2\plus{}4\plus{}8\plus{}15\equal{}29$
$ a_7\equal{}4\plus{}8\plus{}15\plus{}29\equal{}56$
$ a_8\equal{}8\plus{}15\plus{}29\plus{}56\equal{}108$
$ a_9\equal{}15\plus{}29\plus{}56\plus{}108\equal{}208$
$ a_{10}\equal{}29\plus{}56\plus{}108\plus{}208\equal{}401$
$ a_{11}\equal{}56\plus{}108\plus{}208\plus{}401\equal{}773$
$ a_{12}\equal{}108\plus{}208\plus{}401\plus{}773\equal{}1490$
$ a_{13}\equal{}208\plus{}401\plus{}773\plus{}1490\equal{}2872$
$ a_{14}\equal{}401\plus{}773\plus{}1490\plus{}2872\equal{}5536$

There is one way to place 13 stones and 10 ways to place three stones, and no way to place less than three or more than 13 stones.

So the answer is $ a_{14}\minus{}1\minus{}10\equal{}5536\minus{}11\equal{}\boxed{5525}$

Is this too complicated?

Let me reword this into another identical problem:

Consider the number line from 0 to 12. Stones can only be placed integer points. Assuming there is a stone placed on 0, in how many ways can 4 to 12 stones in total be placed on this number line, with the restriction that two adjacent stones cannot be separated by a distance of more than 4?

[hide="Answer"]
It should be 2871. Or am I incorrect?
[/hide]

There has to be an "imaginary stone" (at 13?) at the right hand side to be seperated from by 4 or less...

What I mean is that the river has two banks. You have shown one bank is at 0. Where is the other bank?

Use the generating function

$ (x \plus{} x^2 \plus{} x^3 \plus{} x^4)^n,n\le13$ with the coefficient of $ x^k,k \equal{} 10,11,12,13$.

At least that's the way I attacked that DMAT problem (with a calculator); a non-calculator approach in evaluating the big sum will be appreciated...
(by the way I think 2871 is a bit small, but anyways could you post your work Caelestor?)


Let $a,b,c$ are three lengths of a triangle s.that $1/a+1/b+1/c=3$. Find the minimum of the expression
\[ P=\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b} \]

[quote="greenvert"]$\sqrt{a+b-c}+\sqrt{b+c-a}+\sqrt{c+a-b} \ge \;\frac{9}{\frac{1}{\sqrt{a+b-c}}+\frac{1}{\sqrt{b+c-a}}+\frac{1}{\sqrt{c+a-b}}}\ge \;\frac{9}{\sqrt{3(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b})}}\ge \;\frac{9}{\sqrt{3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}\ge \;3$[/quote]

That is a correct solution. Congratulations, Greenvert! So it is not as hard as Hungkhtn expected it should be. What is your solution, Hung?

[quote="Nameless"][quote="greenvert"]$\;\frac{9}{\sqrt{3(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b})}}\ge \;\frac{9}{\sqrt{3(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}}$[/quote]
Wrong here. $\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ not $\le$  ;)
And hung's problem is still unsolved[/quote]

Oh my eyes!  :oops:  :blush:  :oops:  :blush:  :oops:

[quote="Vasc"]The answer follows immediately from the inequality
$\sum\frac{1}{y^2+z^2} \geq \frac{10}{(x+y+z)^2}$,
where $x,y,z$ are non-negative numbers.
In my book (which will appear in this spring), the following problem there is:
If $x,y,z$ are non-negative numbers and $\frac{-3}{4} \leq r \leq \frac{7}{8}$, then
$\sum\frac{1}{y^2+ryz+z^2} \geq \frac{4(2r+5)}{(r+2)(x+y+z)^2}$.
I conjectured that this inequality is also valid for $r=-1$.[/quote]

Yes. It's my answer. The result is $\frac{2\sqrt{5}}{3}$.

I have no idea!Can you tell me the result,hungkhtn?

keira_khtn, in fact vasc and hung have posted the solution.
You first make the substitution $z = a+b-c$ and so forth, the problem becomes
$\sum \frac{1}{x+y} = \frac{3}{2}$ implies $\sum \sqrt{x} \ge 2\sqrt{\frac{5}{3}}$
or equivalently,
$\sum \frac{1}{x+y} \leq \frac{3}{2}$ implies $\sum \sqrt{x} \ge 2\sqrt{\frac{5}{3}}$

We only need to prove the contrapositive, i.e.
$\sum \sqrt{x} \le 2\sqrt{\frac{5}{3}}$ implies $\sum \frac{1}{x+y} \geq \frac{3}{2}$
or equivalently,
$\sum \sqrt{x} = 2\sqrt{\frac{5}{3}}$ implies $\sum \frac{1}{x+y} \geq \frac{3}{2}$
replace $\sqrt{x}$ by $x$,
$\sum x = 2\sqrt{\frac{5}{3}}$ implies $\sum \frac{1}{x^2+y^2} \geq \frac{3}{2}$

This is vasc's inequality ($\sum \frac{1}{x^2+y^2} \ge \frac{10}{(x+y+z)^2}$) which is posted in the forum before.

[quote="keira_khtn"][quote="Soarer"]We only need to prove the contrapositive, i.e.
$\sum \sqrt{x} \le 2\sqrt{\frac{5}{3}}$ implies $\sum \frac{1}{x+y} \leq \frac{3}{2}$[/quote]
I think you 've reversed the sign.Thank you very much,Soarer![/quote]
Yes you are right, sorry for the mistake... post fixed.


calculus
integration
derivative
algebra
polynomial
calculus computations

Evaluate $ \int_1^e \frac {(x^4 \minus{} 4x^3 \plus{} 9x^2 \minus{} 6x \plus{} 8)e^x}{x^3}\ dx$.

Most of your posts make no sense to other people, just answers.

We need full solutions. (I'm pretty sure this problem was created by thinkin' in "derivatives.")

I'm trying to express it in the form $ \int e^x (f(x) \plus{} f'(x)) dx$  :maybe:  :maybe:

hello, this integral can't be expressed in a closed form by elementary
functions.
Sonnhard.

It doesn't balance out. We have $ \frac{d}{dx}x^{\minus{}2}e^x\equal{}(x^{\minus{}2}\minus{}2x^{\minus{}3})e^x$ and $ \frac{d}{dx}x^{\minus{}1}e^x\equal{}(x^{\minus{}1}\minus{}x^{\minus{}2})e^x$, so $ \frac{2x^2\minus{}2x\minus{}4x\plus{}8}{x^3}e^x$ is the derivative of $ \frac2xe^x\minus{}\frac4{x^2}e^x$. That leaves some polynomial terms we can handle, and an extra $ \frac{7}{x}e^x$. That's well-known not to have an elementary antiderivative, and the answer isn't an elementary number.

This doesn't fit with kunny's established patterns, so it's probably a mistake in the problem.

hello, this is the reason because i used a numerical method to solve this integral.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your integral is $ \int_1^e \frac {(x^4 \minus{} 4x^3 \plus{} 9x^2 \minus{} 6x \plus{} 8)e^x}{x^3}\ dx \approx 28.8941398700195818540$.
Sonnhard.[/quote]

if your're really a person with doctorate title,, you may give us your full solution instead of using your calculator..
thx


function
vector
calculus
derivative
integration
logarithms
advanced fields

Hi everybody!
I was wondering if anybody knows the answer to this:

Suppose you have a function $ f: B^2\to\mathbb R$ which belongs to $ L^1$.

Is it possible to express it as the divergence of a vector field?

In other words, is it true that there exist a $ g\in W^{1,1}(B^2,\mathbb R^2)$ such that $ f\equal{}\partial_1 g^1\plus{}\partial_2 g^2$, where $ g^i$ are the components of the vector $ g$ and $ \partial_i$ are the derivatives (defined in some suitable way, e.g. as distributions).

Any help is appreciated!   :)

Well, if $ f(x,y)$ is your function, consider the vector field $ g\equal{}(F(x,y),0)$ where $ F(x,y)\equal{}\int_{[0,x]}f(x,y)\,dx$ (this integral makes sense a.e. and defines a locally integrable function). You should be somewhat careful when defining the sense in which $ f\equal{}\text{\,div\,}g$, of course, but if all you are looking for is the identity  $ \int \langle g,\nabla\psi\rangle\equal{}\minus{}\int f\psi$ for smooth compactly supported $ \psi$, it holds.

If you really want the $ W^{1,1}$ class, i.e., that all partial derivatives of $ g$ be in $ L^1$, then the answer is "No". Indeed, if $ Q\equal{}[\minus{}1,1]^2$ , then $ W^{1,1}(Q)\ni g\mapsto \text{\,div\,g}\in L^1(Q)$ is a continuous mapping between two Banach spaces. If it were onto, then the image of the unit ball in $ W^{1,1}(Q)$ would contain a small ball in $ L^1(Q)$. Now, consider $ f\equal{}\chi_{[\minus{}a,a]^2}$ with small $ a>0$ and notice that if $ \text{div\,} g\equal{}f$, then $ \|\nabla g\|_{L^1(Q)}$ must be at least about $ a^2\log\frac 1a$ because we have $ \int_{r\mathbb T}\langle g,n\rangle\equal{}\int_{r\mathbb D}\text{\,div\,} g\equal{}a^2$ for all $ 2a<r<1$, so $ g$ must change by at least $ r^{\minus{}1}a^2$ on $ r\mathbb T$ whence the integral of $ \nabla g$ is at least $ a^2\int_{2a}^1\frac{dr}r$.

If I remember it right, the correct statement is $ \text{div\,}: W^{1,1}_{loc}\stackrel{\text{onto}}{\to}H^1_{loc}$ but, frankly speaking, I forgot my PDE too well to guarantee anything here :blush:. Anyway, I hope you've got the idea :).


geometry
circumcircle
trigonometry
trig identities
Law of Sines

I need help on understanding the solution to this problem.

Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C,$ respectively. Then $pq = R^2$ ($R$ is the circumradius)

The solution that book gave was merely:

[hide="Solution"]

$c = 2p \sin B = pb/R, b = 2q \sin C = qc/R$. Multiply and simplify[/hide]

I don't get how they got $c = 2p \sin B$ and $b = 2q \sin C$. I attached a diagram to help people who will help me. Thanks for the help!

Your diagram is incorrect. The two circles go [b]through[/b] A; they aren't centered at A.

Hope this is a right diagram. I still don't see how though.

That's right. Let the centers be $O_1$ and $O_2$ of the circles through $B$ and $C$, respectively.

Since $O_1B = O_1A$ and $O_2C = O_2A$ so both triangles $BO_1A$ and $CO_2A$ are isosceles. Drop the altitudes from $O_1$ and $O_2$ and look at the right triangles formed. Notice that the angles in the right triangles are $B, 90-B$ and $C, 90-C$.

EDIT: Wait, the diagram is still wrong. The circles should be tangent to $BC$ and $B$ and $C$; $AB$ and $AC$ are not the diameters.

Man, I did this one a long time ago....

Well, excuse my poor paint skill

So let the center be $O$ and $P$, respectively, and draw the diameter of each circle through $B$ and $C$, and connect the other end point to $A$ to form two right triangles.  You may not know this, but the angles $B$ and $C$ are actually extending the arc that the other angle of the two right triangles are extending, so they have the same angle measurements (which I have tried to show by the lame marks).  Using law of sines on these two right triangle immediately gives the result.

p.s.: I draw in the circumcircle because I thought it will be useful...(since I believe that section 1.1 is on the law of sines?)

p.s.2: A lot of the problems in Geometry Revisited needs to be solve with the technique they use for the proofs they offer in the section.  Read each proof carefully and understand those techniques will greatly increase your problem solving skills.

p.s.3: has not come out yet :rotfl:


$\text{Find the sum to n terms of the series: }$

$\frac{3}{1^2\cdot 2^2}+ \frac{5}{2^2\cdot 3^2}+ \frac{7}{3^2\cdot 4^2}+\cdots$

[hide] ok i just did it the old fashion way and tested a few values. is it n(n+2)/(n+1)^2?[/hide]

[hide="Hint 1"]Make it telescope.[/hide]

[hide="Hint 2"]$\frac{1}{k^2}-\frac{1}{(k+1)^2}$
What does that equal?[/hide]

as chesspro mentioned, $\frac{2x+1}{x^2(x+1)^2}=\frac{1}{x^2} - \frac{1}{(x+1)^2}$.

so $\displaystyle\sum_{i=1}^{n}\frac{2i+1}{i^2(i+1)^2} =\displaystyle\sum_{i=1}^{n} \frac{1}{i^2} - \frac{1}{(i+1)^2}$. thus the sum telescopes, leaving $1-\frac{1}{(n+1)^2}$


there are unlimited number of 2 squared blocks of the color red
unlmited number of 2 squared blocks of the color blue
unlimited number of 2 squared blocks of color green
unlimited number of 1 square blocks of color black
unlimited number of 1 square blocks of color white

let $a_{n}$ be the number of ways you can pave a straight line of $n$ length  from these blocks

find $a_{1}$
find $a_{2}$

find a general formula for $a_{n}$

[quote="Lena"]thank you, weiquan

can you please explain to me how you arrived at

$a_{n}= 2a_{n-1}+3a_{n-2}$ ?[/quote]

There are 2 ways to extend by one square and 3 ways to extend by two squares.


geometry
trapezoid
circumcircle
geometry proposed

Let $ABC$ be a triangle with $P,Q$ the points of tangency of the incircle with the sides $AC$ and $BC$ respectively. Let the tangency points of the excircles opposite $A$ and $B$ and the line $AB$ be $X$ and $Y$ respectively. If $X,Y,P$ and $Q$ are concyclic, find angle $C$.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=34214 .

  darij

I'm so sorry, but I don't think the solution you are referring to is correct, Darij.
Any isosceles triangle works too, the quadrilateral will then be an isosceles trapezoid (If that is the correct name...  :D  ) and will always be cyclic.

No?

I think the problem with the solution is that Sailor assumes that the triangle is not isosceles, else $K$ doesn't have to be on the circumcircle of $\Delta ABC$, because angle bissector = perpendicular bissector

Hey Jan , that's the geometry problem we saw in wepion the last weekend.
Indeed very nice  :)

Indeed gaincarlo  :D
I'm curious to see the solutions of other people...


calculus derivative Euler articles function conics ellipse

"ibAUIiUduial\uh\ b\idbcaig\io oidgo"

solve this using logic, knowing that each of any sign
 there corresponds to a differente letter or characters as "," , "." etc...

it's written in [b]english[/b] and it was said by a famous mathematician

it's Too much easy

It is from the Famous mathematician from Ancient Greek

$Jesus$  $H.$  $Christ$

He Said

"The Wine Become Water in the same proportion of the water become wine"

So, This is a chemistry problem

Please, someone move this topic to the Chemistry

Thanks

...

stop this, Thales!!
when you play, I play, but this time I'm not playing!

no jokes on this post, please.

are you bullying me!?!?

everybody here bullying me

im still unhappy for all complexes $x$

neether the case $x=0$

and others

stop this, thales. nobody here is bullying you.
I just think that there is time for jokes, and time for serious math.

this topic should be serious man, please

to put it less controversially,

thales418 ~ Fabiomr1

Thales

im Thales because of the great mathematician of ancient greek

if i made another account it would be:

Galois

or something else

418 is not a common number

google it

i challenge fabio to write the solution of this problem using only the number 3

i will do the same using just the number 7

lets start

Hey I am confused. Was the famous Greek mathematician Thales or Archimedes?

damm!!! I'm not thales!!

won't any of you try the sequence?

hmm...

this is the fibo sequence!!

and i win!

That is NOT the Fibo sequence. This is: 0 1 1 2 ...  :roll:
How about calling it the Fabio sequence?

hey Buda

lets call this  "Thales-Fabio-Buda Sequence"

:)

hey Thales let's add Feynman's name onto it and right something sort of like
in
http://math.ucr.edu/home/baez/crackpot.html
to write how it explains major scientific laws

hey mister function

are you stealing my conjecture?!?!

too bad....

make your own conjecture and dont prove it, never prove a conjecture made by yourself!!!

in one day, some idiots will prove this for you, in 5 years, or in 350...

he

Thank you very much, but I really don't deserve the honour of having a derivative of my name attached to this ground-breaking work.

By the way, which is defective: your keyboard or your eyes?

Oh, and as regards the above post:
Poincare made the Poincare conjecture in 1896.
Poincare proved the Poincare conjecture in 1900.
Poincare gave a counterexample to disprove his earlier proof of the Poincare conjecture in 1904.
There is a school of thought of the opinion that Poincare's proof was current, he only thought in 1904 that the other fools should work it out for themselves ...

hello both mister tall guy
and Buda

i just wanted to point out 418 is not too special

oh and

no

i already have my own

stuff

it's called the misterfunction sequence

it actually

explains both mysteries in ecology

and is really quite linked to pascal

however i chewed
up my proof

Hey Function-Buda

the Thales-Fibo-Fabio-Euler Sequence is a unknow sequence, yet :( :(

but, according to the prophecy, this sequence has the True of the Trues!!

a true so big, that this value is also infinity!!

i will pay 500 dollars to the first who gave me this formula

pie!?!?

sorry, im not hungry

 can i choose a cake?

or only a juice

ho ho ho

nobody solved my puzzle

I will double your bet, Thales; I'm now giving away [b]1000[/b] dollars to who solve my puzzle

I will give $1000+n$ Dollars to the first who solve this amazing pluzze!!

i will give an infinitesimal number of euros to the winner....
well than everyone can see it has to do with euler

hey function

o will integrate you if you continue doing this idiot things

this is a very very polite and formal topic

dont joke!

[quote="funcia"]ibAUIiUduial\uh\ b\idbcaig\io oidgo
for every letter put a number and you will get pi[/quote]

That's wrong! I tried it out and got phi (withe an extra h, see?)

oh yes i see. i didn't try it i just read an article on the such on my homepage(mathworld  :P )

lets close this topic

the solution is trivial, the solution is $3$  or pi, or sometimes called phi

and by anothers phhhhhhhh.....hhi

becaseu $h=0$

the solution is Pi

yeah well it is the number
Thales-Fibo-Fabio-Euler Sequence
TFFES

if we draw a circle, a ellipse, a square, a line, a triangle, a n-gon and finally Revolution Hyperboloid

the intersection of these 3 figures is a point

i will give 665,99 dollars for the first ho prove this

where do we draw them? 500 miles away from each other?
i offer arc tan 6 yen for whoever tells me why i am an idiot in this matter

hey function

you is undefined

you does not exist

so, stop talking


Let $a,b,c>0$ such that $a+b+c\geq\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$.Prove that:

     $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq\frac{3}{2}$ ;)

What you are saying that the inequality that I posted it's not true? :?

Yes if you have another opinion put a solution

[quote="silouan"]This ineq holds if and only if $ a+b+c\leq3 $ . But from the condition we have that $ a+b+c\geq3 $ . So you should turn the ineq.[/quote]

I think it does hold, take a=b=c=2, which is the case of a+b+c>=3, and you can easily see that the inequality still holds..  :?

I'm trying to use Cauchy, but trying to do it in my head only makes it a lot harder. I'll reply when I find pen and paper, and the proof. :D

Of course, this inequality does hold. :D

I suppose that $a+b+c\leq3$.
We have  $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq\frac{9}{3+a+b+c}\geq\frac{3}{2} $

I prove that the ineq holdw for $a+b+c\leq 3$ not for $a+b+c\geq3$.
It could be right for $a+b+c\geq3$ but I found it only $a+b+c\leq 3$

[quote="silouan"]I suppose that $a+b+c\leq3$.
We have  $\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq\frac{9}{3+a+b+c} $[/quote]

but why is this true?

Silouan, you can't suppose that $a+b+c\leq 3$ :rotfl:
 Man, you are unbelievieble. :D

Sorry big mistake I have in my mind another ineq. :oops:  :blush:

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:
Sorry, but this IS pretty funny.

Silouan, what is this inequality you have in mind?

The desired inequality
$\displaystyle \frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq\frac{3}{2}$
is equivalent to
$\displaystyle \frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\leq\frac{3}{2}$,
or
$3abc+ab+bc+ca \ge 3+a+b+c$.

It suffices to show that $abc \ge 1$ and $ab+bc+ca \ge a+b+c$.
We have (by AM-GM ineq)
$3(a+b+c) \ge (2a/b+b/c)+(2b/c+c/a)+(2c/a+a/b) \ge 3 \frac{a}{\sqrt[3]{abc}}+3\frac{b}{\sqrt[3]{abc}}+3\frac{c}{\sqrt[3]{abc}}$,
hence $abc \ge 1$.
Also (by Cauchy-Schwarz),
$a+b+c \ge \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \ge \frac{(a+b+c)^2}{ab+bc+ca}$,
hence $ab+bc+ca \ge a+b+c$.

Finally, someone has done it! Anyway Vasc's solution is almost like my solution with a slight difference. ;)


Posting  without comments - the result of a recent poll:
[img]http://img.photobucket.com/albums/v202/chowk_user/feelingsamericans.gif[/img]

I think the real question here Gyan, is what does your dog think about the election?  ;)  :D

It seems that the Philipines, Poland and India want Bush more even than the Americans :D China doesn't care, and everybody else thinks it's a bad thing ...

A few things, numbered for my own amusement:

(1) Margin of error is one of the things least likely to mess up poll results.  The reason you hear so much about it is that it is relatively easy to quantify.
(2) Even if there were, say, a margin of error of 10 points (which is really, really large), it would allow for at most one country in addition to those named (South Africa) to be within the margin of error.
(3) I don't think it's reasonable to include Poland in that list, any more than it would be to say, "Wow, those Japanese really hate us -- only 6% think better of us now."  They really both belong in the class of countries that overwhelmingly did not change their opinions of Americans.
(4) This survey misses two very important groups of people: those who assumed we would do something bad and haven't had their opinion of us change as a result of electing Bush (but who would be on the left if Kerry were elected) and those who assumed we would do something good and haven't had their opinion of us change as a result of electing Bush (but who would be on the right if Kerry were elected).
(5) I wonder what everyone in between India and Turkey thinks.  Also, the Australians and Ukranians.

[quote="MithsApprentice"]Now let's find out the margin of error for those polls.[/quote]
The, Poll was conducted by  [url=http://news.bbc.co.uk/2/hi/americas/4185205.stm]BBC World Service[/url] and they claim the margin or error is between 2.5 and 4 % (depending on the country).

Poll was conducted over last few months,


Sa se arate ca ecuatia: y la puterea 4 adunat cu x la puterea 3 =7 nu are solutii naturale.

Sa se arate ca ecuatia: y^{4}+X^{3}=7 nu are solutii naturale.

Se "rade" totul modulo 11 daca nu ma insel.

Nu cumva enuntul original spune ca ecuatia nu are solutii [b]intregi[/b]?
In numere naturale problema este prea simpla.

Ceva asemanator...
  Sa se arate ca ecuatia $\ x^2-y^3=7$ nu are solutii intregi.

Eu am facut-o in alt mod...Uite solutia mea:
 Daca $\ y$ este par, atunci avem $\ x^{2}=8k+7$, care nu are solutii.
 Daca $\ y$ este impar, relati se rescrie astfel:$\ x^{2}+1=y^{3}+8$, sau $\ x^{2}+1=(y+2)(y^{2}-2y+4)$, de unde folosind ca $\ y=2k+1$, ajugem cu usurinta la $\ x^{2}+1=4(k+1)(4k^{2}+3)$, si avem $\ 4k^{2}+3\mid x^{2}+1$, ceea ce e imposibil caci $\ 4k^{2}+3$ nu poate fi o suma de patrate, deorece $\ (2k)^{2}$ este patrat perfect, in timp ce $\ 3$ nu e. :wink:

se pot crea o infinitate de exercitii de genul acesta

problema asta a fost descoperita de Lebesgue.


Let $ p$ be a prime, let $ k$ be a number not divisible by $ p$, and let $ b$ be a number that has a $ k^{th}$ root modulo $ p$. Find a formula for the number of $ k^{th}$ roots of $ b$ modulo $ p$ and prove that your formula is correct.

Reduce it to http://www.mathlinks.ro/viewtopic.php?t=192756 (it's just one step to do). Or take some search.


Hi, I need some help about a problem -- irrationality of e^n.
[hide]proof that for all n(n is natural number) e^n is irrational[/hide]

We know Fourier's proof about e. (ofcourse we can do this with transcendent number concept)
And I applicate it to (to solve problem for) e^n. But I can't get anything.
Can anyone help me about it?
Can anyone suggest any sollution?

Thanks..

This is a famous problem with multiple published solutions, none of which should be detailed on the High School Basics board.  College playground, maybe.

Use the Taylor series approximation for $ e^{x}$.  That is,
$ e^{x}\equal{}1\plus{}x\plus{}\frac{x^{2}}{2!}\plus{}\frac{x^{3}}{3!}\plus{}\cdots\plus{}\frac{x^{n}}{n!}\plus{}R_{n}(x)$,
where $ R_{n}(x)$ is the remainder term, with $ R_{n}\leq\frac{3}{(n\plus{}1)!}$.  Assume that $ e^{x}$ is rational, so $ e^{x}\equal{}\frac{a}{b}$.  Now choose $ n$ sufficiently large ($ n>b,3$).  Then multiply through by $ n!$.  What happens?  What can you conclude?


calculus
integration
inequalities unsolved
inequalities

Let $n$ be a positive integer, $n \ge 2$.  Prove that
\[\sum_{k = n}^{2n - 2} \frac{1}{k^2} < \frac{1}{2n - 1}.\]

Thanks, Darij!  I tried comparing to the integral, but it wasn't quite working.


I came up with a stupid inequality,
show that,
$\sqrt{ \frac{1}{1990}+\frac{2}{1989}+...+\frac{995}{996}}<\frac{1990}{4}$


AMC
AMC 12
AMC 12 A
USA(J)MO
USAMO
HCSSiM
AIME

Cool, thanks!

edit: neato, I'm the top AMC12A scorer in Ohio :p

lol i got top amc 10 b and like top 30ish for 12 a...

thanks for the post

Lol I got top 3 in Ohio for the 10B.  I'm the 9th grader.

Are there any "global" statistics. Like total taken. Average by gender etc...?

Yes, we will post global statistics when they become available.  They are not yet compiled.

Steve Dunbar
MAA Director, American Mathematics Competitions

No, top 9 in NJ in AMC 10. I was hoping for like top 3...and nobody in NJ got a 150 on either test. :o

Lol. I got top AMC 10A and top 6 for AMC 12B.  In... my city. =P lol.

sniffle top 10  on both 10A, and 10B. wasn't realy sure about 10B cuz there was a 3 way tie for 126. in 8th grade, i was 3rd on 10A after mathcrazed and some other kid in his school.

If I had taken the 10A in Ohio (I took in MI), I would've been 2nd.
If I had taken the 12B in Ohio (once again, MI), I would've been...tied for 17th :D.

In MI, I was tied for 5th for 10A :(
I was tied for 14th on 12B :(

I feel sad now.

Tied for 13th in CA for the 10B.
Tied for 31st in CA for 10A.
I could have done much better though...

Lawl, same as Carbon57 (13th) in CA for the 10B, even though I kinda failed it.
Eh, like a billionth on the AMC 12A...whatever, I was sick.  :P

yes! i got really high on the B in CA

my friend got 2nd place CA...

Do we get prizes for being in top 3 state overall?

I got best AMC 10a score in my tiny, minuscule state.

Hm, 4th on the 12A, 2nd on 10B. Not bad. I can guess who beat me on the 12A and know who did on the 10B.

[quote="123456789"]Wait, if I'm school winner for 2008-2009 AMC 12 A and B, and also 2009-2010 AMC 12 A and B, do I get considered school winner twice or four times, thus earning a gold medal?[/quote]

nope.  The medals are given for winning in consecutive [b]years[/b] NOT tests.

[quote="serialk11r"][quote="Alumen"][quote="serialk11r"]Heh 3 145.5 freshmen are Alumen, Databox, and Quatto. 150 must be CircleSquared, nice job.
I'm one of those 6 freshmen tied for 30somethingth :([/quote]
Isn't Circle a 8th grader though?
Or maybe I'm just misinformed.[/quote]
Yes Circle is Patrick, Harker kid who got 2nd at State last year, 8th grade now. Argh I should be up there.[/quote]

But if Circle is an 8th grader, then he didn't perfect the 10A, since a freshman got a perfect on it according to the state stats...

Yeah, shocked about my placement.

I got first in 10B and 12A with 144 and 138 respectively in Michigan.

hm nobody got 145.5 or 150 for AMC10A for NJ, so I'm one of the 3 that got the highest AMC 10A scores...
It's also disappointing because if I got 150 then I would be unique...same with the other 2...if any of them got 150 then he would be unique...

[quote="123456789"]Wait, if I'm school winner for 2008-2009 AMC 12 A and B, and also 2009-2010 AMC 12 A and B, do I get considered school winner twice or four times, thus earning a gold medal?[/quote]

From my understanding, you'd get 2 bronze medals for that. I'm not entirely sure though.

[quote="serialk11r"][quote="Alumen"][quote="serialk11r"]Heh 3 145.5 freshmen are Alumen, Databox, and Quatto. 150 must be CircleSquared, nice job.
I'm one of those 6 freshmen tied for 30somethingth :([/quote]
Isn't Circle a 8th grader though?
Or maybe I'm just misinformed.[/quote]
Yes Circle is Patrick, Harker kid who got 2nd at State last year, 8th grade now. Argh I should be up there.[/quote]
I know. I should be tied for first too.

(Jk, in reality I wasn't anywhere close to solving that problem.... ehh whatever :P)

Looking at all of the stats, I thinks I was th only 7th grader or below to get a 150 on the AMC10. :P

First place 10A in NV!  Tied for 5th 12B.

1st in IN on the 12A
3rd in IN on the 12B to my two teammates
Our team was first and second in IN (1-B, 2-A)
as well as first and second in the region


this is a very cool problem:

a)Find all positive integers k  such that there exists two positive integers  a  and  b greater than k such that  (a-k) , (b-k) and (ab-k) are all perfect squares.

b) Find all positive integers k  such that there exists two pisitive integers a and b greater than k such that (a-k) ,(b-k) and (a+b-k) are all perfect squares.

i have a solution for a) so if no one solves it ill post it. i haven't really worked on b) :D  :P

enjoy¡¡¡ :lol:

nobody??? it's really not that hard¡¡¡¡ :)

yes i guess you can  ;)  sucks cause it took me like an hour to find my a´s and b´s  when the answer is so simple. :mad:  anyway the really hard part is b) :P


geometry
3D geometry
algebra
polynomial
Vieta
quadratics
complex numbers

$a,b,c$ are complex numbers such that $a+b+c=2$, $a^2+b^2+c^2 = 3$, and $abc=4$. Then compute $a^3+b^3+c^3$.

BONUS: If anyone can solve this problem by actually finding $a,b,c$, I'll be very surprised because I don't know how to find $a,b,c$. Although I know how to find $a^3+b^3+c^3$.

Good luck.

don't know how to find a, b, c individually at the moment but my solution is as followed...

[hide]we can start out by finding ab+bc+ac as...
$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$

substitute given information...

$ab+bc+ac=\frac{1}{2}$

the second equation i used...
$a^3+b^3+c^3=(a+b+c)^3-3(a+b)(b+c)(a+c)$

we can manipulate the equation $a+b+c=2$ slightly to get

$a^3+b^3+c^3=(a+b+c)^3-3(2-c)(2-a)(2-b)$

multiplying out, we get $a^3+b^3+c^3=(a+b+c)^3-3(8-4(a+b+c)+2(ac+ab+bc)-abc)$

substitute given information

$a^3+b^3+c^3=(2)^3-3(8-8+1-4)$

then we get $a^3+b^3+c^3=17$

i hope thats correct. i spent a lot of time (probably too much time)[/hide]

I have a different solution. I'm not very good with tex, it will take me hours, but I'll outline my solution and everyone can just work it out for themselves. Mine might be a bit tedious.

[hide] Call a+b+c the "first one" a^2+b^2+c^2 the "second one" and abc the "third one". Cube the first one, and multiply the first one by the second one, creating two equations. Subsitute the third one in wherever needed. You now have a nice system of equations. Cancels out nicely with adding. I got 17[/hide]

[hide]
Use [url=http://mathworld.wolfram.com/VietasFormulas.html]vietas theorem[/url]:
Suppose $a,b,c$ are the roots of an equation $x^3 + a_1x^2 + a_2x + a_3^3$
We are given $a + b + c = 2$ thus $a_1 = -2$ by definition.
We are given $abc = 4$ thus $a_3 = -4$ by definition.
We are not given $ab + bc + ac$ however that can be deduced using the given information:
\[(a+b+c)^2 = a^2 + b^2 + c^2 +2(ab +bc + ac)\]
\[4 - 3 = 2(ab + bc + ac) \implies \displaystyle \frac{1}{2} = ab + bc + ac\]
therefore $\displaystyle a_2 = \frac{1}{2}$
So we need to find the roots of the eqauation $x^3 -2x^2 + \frac{1}{2}x - 4$
And well, there's two complex roots, so I plugged it into maple and it's 17.

But yeah, solving it the other way is a lot cleaner.

btw, this is what maple returns:
[code]
2.458440227, -.2292201136+1.254793225*I, -.2292201136-1.254793225*I
[/code]
[code]

[/hide]

Yeah, you can generalize it to any polynomial though. I just edited the post to include a link to the math world definition.

I fixed a typing error too :P. there was an $a_2$ I put that was suppose to be $a_3$

there is a way to find a, b, and c
[hide]i let the complex numbers be the roots of $x^3+bx^2+cx+d=0$ with the information provided, i created newton sums:
($s_1$)+b+0
($s_2$)+b($s_1$)+2c=0
($s_3$)+b($s_2$)+c($s_1$)+3d=0
$s_1$ (sum of roots) and $s_2$ (sum of squares of roots) were given as well as d (since product of the roots is the constant term)
this allowed you to solve for $s_3$ (sum of cubes of roots). $s_3$=17
we also can create the polynomial from which the roots came from.
$x^3-2x^2+\frac{1}{2}x-4=0$
thus a, b, and c can be found. i just don't feel like doing that now[/hide]
aw, somebody came up with the polynomial while i was typing :(

[quote="Silverfalcon"]That's really interesting way. :D

I'm still bit confused on how Viete's formula is applied here though. Wasn't Viete's formula like $\frac{-b}{a}$ equals the sum of the roots of quadratic and $\frac{c}{a}$ equals the product of the roots of the quadratic?

Or am I missing something...?[/quote]

it depends on the # of terms. if u take a look at mathworld for a sec, you'll notice that in the 3rd degree polynomial, the product of the roots is -(c/a), as opposed as in a 2nd degree polynomial, the product is (c/a)