function
integration
calculus
real analysis
real analysis unsolved

Let the function $ f$ have the property

$ f(x)\plus{}f^{\minus{}1}(x)\equal{}x$ for all $ x\in\mathbb{R}$

then find $ \int^{a}_{\minus{}a}f(x)dx$

$ x  \minus{}> f(x)$: ....   $ f(f(f(x)))\equal{}\minus{}x$ and $ x  \minus{}>  f(x)$ have $ f(\minus{}x)\equal{}\minus{}f(x)$ and $ \int^{a}_{\minus{}a}f(x)dx\equal{}0$

Er, show your work?

I made it through your first three dots but I certainly don't find obvious the conclusion that $ f$ is an odd function.

FWIW, one function with the required property is f(x)=Cx. In this case, we have C+1/C=1, and C=(1+sqrt(3)i)/2 or (1-sqrt(3)i)/2.
In both cases the intergral is zero.

Edit: fixed per LydianRain's observation.

In both cases $ f$ integrates to $ 0$, since it is an odd function.


calculus
integration
function
algorithm
quadratics
algebra
rational function

hmm I kind of want to say to use Trigonometric Substitution, try to get it to fit [img]http://upload.wikimedia.org/math/a/1/e/a1e54d721e1fea9d5a641cda2944ca34.png[/img]

as a start.....

[quote="trichoplax"]The substitution x = (t - 1/t)*sqrt(A)/2 converts the integrand to a rational function of t, which is good enough for me.[/quote]
I think it's equivalent tu sabstitute $sh(x)$. If you solved it then put full sulution

[quote="Extremal"]
I think it's equivalent tu sabstitute $sh(x)$. If you solved it then put full sulution[/quote]
I did not continue. Since there is a well-known algorithm for integrating any rational function, it is my opinion that an integral is fully solved once it has been reduced to that case. Mathematica's solution is relatively short, so it must have used a smarter substitution than I did. Maple's solution is a mess.

[quote]Since there is a well-known algorithm for integrating any rational function, ... [/quote]
That comes with one "but." There is an algorithm, provided you can factor the denominator into linear and quadratic factors. The existence of that factorization is assured by the Fundamental Theorem of Algebra, but the FTA is a non-algorithmic pure existence proof. There is no universal factoring algorithm, and in fact some kinds of factoring algorithm would be prohibited by the Abel-Galois non-solvability results.

Of course in this case, the denominator will come in a factored form; that particular objection will not apply here.

[quote="Extremal"]find $\int\frac{\sqrt{x^{2}+A}}{x^{2}+B}dx=?$ where $A,B$ is real.[/quote]
Ok, My answer is next:
$\int\frac{\sqrt{x^{2}+A}}{x^{2}+B}$
subtitute $\sqrt{A}sh(y)=x$ then we have
$\frac{\sqrt{A}}{A}\int\frac{ch^{2}(y)}{ch^{2}(y)-1+\frac{B}{A}}$
then we put $ch(y)=\frac{e^{y}+e^{-y}}{2}$ and  di new substtitute
$e^{2x}=z$ It's done.


Consider the positive integers that such that its digits are either $ 1$, $ 2$, or $ 3$.
Define a function $ f(n)$ such that f(N) = the number of positive integers consisting of only $ 1$, $ 2$, and $ 3$ whose digits sum to $ N$.

Prove that $ f(N \plus{} 3) \equal{} f(N \plus{} 2) \plus{} f(N \plus{} 1) \plus{} f(N)$.

I'm confused.

So do you mean that (ii) is a specific case of the recursion at the bottom? Do you want us to prove the recursion, and then use that to find $ f(10)$? If so:

[hide="Abridged Solution"]This is the same as the number of ways of traversing a series of stepping stones, where from each stone you may move to the next, or skip either 1 or 2 stones.

The number of ways to get to stone $ N \plus{} 3$ is thus the sum of the number of ways to get to stones $ N \plus{} 2$, $ N \plus{} 1$, and $ N$, since these are the possible locations immediately preceeding stone $ N \plus{} 3$. This gives $ f(N \plus{} 3) \equal{} f(N \plus{} 2) \plus{} f(N \plus{} 1) \plus{} f(N)$. The initial conditions are $ f(1) \equal{} 1$, $ f(2) \equal{} 2$, and $ f(3) \equal{} 4$.[/hide]
[hide="Answer"]Using the recursion and the initial conditions:
$ \begin{array}{ccc} f(4) & \equal{} & 7 \\
f(5) & \equal{} & 13 \\
f(6) & \equal{} & 24 \\
f(7) & \equal{} & 44 \\
f(8) & \equal{} & 81 \\
f(9) & \equal{} & 149 \\
f(10) & \equal{} & 274 \end{array}$[/hide]

[quote="unimpossible"]Well its a correspondence because  the numbers inf(n+3) can be rewritten as either a number in f(n) with a 3 attached at the end, a number in f(n+1) with a 2 attached at the end, or a number in f(n+2) with a 1 attached at the end.[/quote]

To be slightly more precise:  let $ \mathcal{F}_n$ denote the set of positive integers consisting of $ 1$s, $ 2$s, and $ 3$s whose digits sum to $ N$.  Then $ f(n) \equal{} |\mathcal{F}_n|$ and the above argument can be made precisely by considering the last digit of any element of $ \mathcal{F}_{n \plus{} 3}$.

[hide="Fun stuff"] Note that the generating function for this sequence is given by $ \frac {x \plus{} x^2 \plus{} x^3}{1 \minus{} (x \plus{} x^2 \plus{} x^3)}$, which can be used to prove that its characteristic polynomial is $ x^3 \equal{} x^2 \plus{} x \plus{} 1$ as desired.
[hide="In particular..."] Let the three roots of $ x^3 \equal{} x^2 \plus{} x \plus{} 1$ be $ \alpha, \beta, \gamma$.  Then

$ \frac{x \plus{} x^2 \plus{} x^3}{1 \minus{} (x \plus{} x^2 \plus{} x^3)} \equal{} \frac{x \plus{} x^2 \plus{} x^3}{(1 \minus{} \alpha x)(1 \minus{} \beta x)(1 \minus{} \gamma x)}$.

Partial fraction decomposition will show you that $ f(n) \equal{} a \alpha^n \plus{} b \beta^n \plus{} c \gamma^n$ for suitably chosen $ a, b, c$, from which the conclusion follows (in a much more roundabout way than the simple counting argument).  I include this argument merely to demonstrate the general power of the generating functions approach. :) [/hide] [/hide]


x^5+31=y^2 show that there is no solution in integers.


inequalities
function
inequalities proposed

$x,y,z \ge 0$ and $x+y+z=1$. Prove that
$\frac{10}{27}+\sqrt[3]{\frac{1}{9}}\le xyz+xy+yz+zx+\sqrt[3]{x^{3}+y^{3}+z^{3}}\le 1$
P/s: To Albanian Eagle: I had edited it!  :lol:

It's very easy at all :). Denote $x=ab+bc+ca+abc$ then the expression is exactly \[x+\sqrt[3]{1-3x}\] then examine this function of $x$, notice that $0\le x \le 10/27$.

To NkMAsTeR. You need to change in your problem $x,y,z$ to $a,b,c$.

Oh, I'm sorry. This is a type error! :D  :blush:
It don't think $1-3x=x^{3}+y^{3}+z^{3}$ because $1-x^{3}-y^{3}-z^{3}=3(xy+yz+zx-xyz)$
But if i replace 1 by 2, can you find the minimum and maximum of this?

just to clearify the problem NkMAsTeR are the variables a,b,c or x,y,z in your problem ??? :P

An open question: Find the best constant $k$ such that
$xyz+xy+yz+zx+\sqrt[k]{x^{3}+y^{3}+z^{3}}\le 1$

[quote="zhaobin"]to explain $Flove$'s idea: \[1-3(ab+bc+ca+abc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)+abc)=a^{3}+b^{3}+c^{3}\] [/quote]
If you like,change $a,b,c$ to $x,y,z$ here,I think you will understand...0.0

[quote="zhaobin"][quote="zhaobin"]to explain $Flove$'s idea: \[1-3(ab+bc+caabc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)+abc)=a^{3}+b^{3}+c^{3}\] [/quote]
If you like,change $a,b,c$ to $x,y,z$ here,I think you will understand...0.0[/quote]
It must be \[1-3(ab+bc+ca-abc)=(a+b+c)^{3}-3((ab+bc+ca)(a+b+c)-abc)=a^{3}+b^{3}+c^{3}\] :lol:

But I wonder how can it solve?  :P Normal way is seemingly impossible,

But with pvthuan, it is only average!So, I won't post it here! :lol:

[quote="Flove"] \[x^{3}+y^{3}+z^{3}-3xyz=x^{2}+y^{2}+z^{2}-(xy+yz+zx)=1-3(xy+yz+zx)\] Let $t=xyz,r=xy+yz+zx$, so \[f(t)=t+r+\sqrt[3]{1+3t-3r}\] Because $r^{2}\ge 3t$ and $r\ge 9t$ so \[1+3t-3r \le 1-3(t+r)+3(t+r)^{2}-(t+r)^{3}\]  \[\Leftrightarrow 6t+(t+r)^{3}\le 3(t+r)^{2}\] Which is obviously true because \[(t+r)^{2}\ge (t+r)^{3}\mbox{and}(t+r)^{2}\ge y^{2}\ge 3t.\] I think the rest is also easy

To NkMAsTeR: [hide]What do you feel about Hungkhtn's book?[/hide][/quote]
But It's not complete! There are two inequalities here! :D
To Flove [hide]I don't like it much! Sorry to you and Hungkhtn[/hide]

Why it's incomplete? I don't understant? The right-hand inequality is ok now, but I think the left-hand inequality is not hard. Anyone could be try? I don't have so much time to enjoy it, because this problem is not totally beautiful.

[quote="NkMAsTeR"]
To Flove:
I don't like it much! Sorry to you and Hungkhtn[/quote]

Oh, in fact it is not as good as hungkhtn marketed it loudly. The best chapter is chapter 3. People and hung talked too much about it. The first two chapters look like a mess and may give readers the feeling that its a collections of problems and solutions.  Many of them are not natural, problems were not arranged in a logical and systematic way. Still too many errors in typos, and writings.

[quote="NkMAsTeR"]Yes, of course it is not beautiful at all, but I think it make me thinking, not just applling an "algorithm".Do you know why I don't like Hungkhtn's book much? Because It has some mistakes and somewhere, the solutions are not creative!
And can you prove the open question of me?[/quote]

To NkMAsTeR: Sometime, generalising a problem is not so important, also finding the best constant are mechanical. I'm very glad to study which part of Hungkhtn's book is as not creative as you say? Chapter II is the chapter where almost problem is the most creative, but in chapter III, the purpose is "Methods". Problems exposing this method are just important.

Let $m=xy+yz+zx$ and $t=xyz$ We have:
$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y)(y+z)(z+x)$
$=1-3m+3t$
$f(x,y,z)=xy+yz+zx+xyz+\sqrt[3]{x^{3}+y^{3}+z^{3}}=m+t+\sqrt[3]{1-3m+3t}=g(t)$ but $m$ is fixed$\Rightarrow$
$g(t)$ get the maximum or minimum when
    1 in 3 numbers $x,y,z=0$
or 2in 3 numbers equa to each other
Amssume $x\leq y\leq z$ and let $x+y=r$
Now we have
$f(0,r,z) =rz+\sqrt[3]{r^{3}+z^{3}}=rz+\sqrt[3]{1-3rz}$ ( Because $r+z=1$ )
Let $rz=a \text{ (}0\leq a\leq \frac{1}{4}$
$\Rightarrow f(0,y,z)=a+\sqrt[3]{1-3a}=F(a)$
 $F'(a)=1-\frac{1}{ \sqrt[3]{(1-3a)^{2}}}=0 \Leftrightarrow a=0$
 Easy to get $F(a)\leq 1$ Because $0\leq a\leq \frac{1}{4}$
$\Rightarrow$ max$f(x,y,z) =1$   :lol:
similaly min$f(x,y,z)= \frac{10}{27}+\sqrt[3]{ \frac{1}{9}}$  :lol:
What do you think? :oops:

Yes, but there are, may be you don't agree with me, some problems solved by S.O.S , are not creative. I'm sorry, but I think, it is really a good book, but, of course, there isn't any perfect things. And, I'm jsut a kid, so, please fogive me if I said somethings wrong.

To Do Tien Dang
[quote="Do Tien Dang"]Let $m=xy+yz+zx$ and $t=xyz$ We have:
$x^{3}+y^{3}+z^{3}=(x+y+z)^{3}-3(x+y)(y+z)(z+x)$
$=1-3m+3t$
[/quote]
I'm sorry, It must be $=1-3m+6t$ . Please check it and sorry if I wrong!

Actually, I don't really like problem using SOS method simply, but in almost case, SOS is the best for three-variable problem (at least better than mixing variables). If you say that SOS is not creative, then you will agree that mixing variables and almost other parts (involving EV, hafl-convex function, generation of Schur...) are also not creative, because your work is only applying, applying and applying. But this only shows that you're just a child.  :lol: Such post as yours is only acceptable is diendantoanhoc, not in mathlinks.  :lol:

As a matter of fact, sometime SOS is mechanical. But there's many extremely hard and nice problem where SOS is unique, but I say it not to negate other methods because I myself don't very like SOS. Such idea shouldn't post here, it should be placed in

http://www.mathlinks.ro/Forum/viewtopic.php?p=630199#630199

To you, NkMAster and other: If you fell that SOS is boring, the only reason is that, you are so professional about it, or you know nothing.  :arrow:

Stop, stop :!:

To Vietnamese boys: It is not uncommon for people to have different points of view. We need arguments and discussions in a polite way. Otherwise, foreign friends will hate you.

Yes, I agree with you. Sorry, Flove.
To pvthuan: Is is inequality okie?


[b]English Version.[/b]

Let $a,b,c$ be non-negative real numbers. Prove that

\[\sqrt{\frac{a}{b+3c}}+\sqrt{\frac{b}{c+3a}}+\sqrt{\frac{c}{a+3b}}\ge\frac{3}{2}.\]

I have a very nice solution to this one.

:lol: So excellent!!! You solution is totally the same as mine!!! Contradiction method is the main key !!! Also, it is the key for another problem of mine here

http://www.mathlinks.ro/Forum/viewtopic.php?t=152337

However, no one finds this key! Thank you very much, Arqady, since all the solutions I know is very terribly complicated.

[quote="arqady"]I have applied this method here too:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64122[/quote]

Actually, I also have the same idea as yours in proving the above inequality. Moreover, I think the following similar one is true, Arqady

\[ \sqrt{\frac{a}{a+2b+c}}+\sqrt{\frac{b}{b+2c+a}}+\sqrt{\frac{c}{c+2a+b}}\le \frac{3}{2}.\]

[quote="arqady"][quote="hungkhtn"]I think the following similar one is true, Arqady
\[ \sqrt{\frac{a}{a+2b+c}}+\sqrt{\frac{b}{b+2c+a}}+\sqrt{\frac{c}{c+2a+b}}\le \frac{3}{2}. \]
[/quote]
I have checked it and I agree with you.
By the way, $ \sqrt{\frac{a}{a+7b+c}}+\sqrt{\frac{b}{b+7c+a}}+\sqrt{\frac{c}{c+7a+b}}\geq1$ already,
but $ \sqrt{\frac{a}{a+5b+c}}+\sqrt{\frac{b}{b+5c+a}}+\sqrt{\frac{c}{c+5a+b}}\le\frac{3}{\sqrt7}$ also. :wink:[/quote]

Yes, I use the same method as above. But it is really complicated! So I really curious of how could you find out fastly two above inequalities?  :maybe: Are your solutions easier?

I am curious too :D How can solve it? It is good to see your solution :)

I have found mistake in the proof of $ \sqrt{\frac{a}{a+5b+c}}+\sqrt{\frac{b}{b+5c+a}}+\sqrt{\frac{c}{c+5a+b}}\le\frac{3}{\sqrt7}.$
The inequality is not true.
Thank you, hungkhtn and  zaizai-hoang.
Probably, $ \sqrt{\frac{a}{a+3b+c}}+\sqrt{\frac{b}{b+3c+a}}+\sqrt{\frac{c}{c+3a+b}}\le\frac{3}{\sqrt5}$ is true,
 but if it's true, it's not easy. :wink:

[quote="arqady"]Let $ \frac{a}{a+\alpha b+c}=\frac{x^{2}}{\alpha+2},$ $ \frac{b}{b+\alpha c+a}=\frac{y^{2}}{\alpha+2}$ and $ \frac{c}{c+\alpha a+b}=\frac{z^{2}}{\alpha+2},$
where $ \alpha,$ $ x,$ $ y$ and $ z$ are non-negative numbers.
Then $ \begin{cases}(x^{2}-\alpha-2)a+\alpha x^{2}b+x^{2}c=0\\ y^{2}a+(y^{2}-\alpha-2)b+\alpha y^{2}c=0\\\alpha z^{2}a+z^{2}b+(z^{2}-\alpha-2)c=0\end{cases}.$
Thus, $ \det\left( \begin{array}{ccc}x^{2}-\alpha-2 & \alpha x^{2}& x^{2}\\ y^{2}& y^{2}-\alpha-2 & \alpha y^{2}\\ \alpha z^{2}& z^{2}& z^{2}-\alpha-2 \end{array}\right)=0.$
Id est, $ (\alpha+2)^{2}=(\alpha-1)^{2}x^{2}y^{2}z^{2}+(\alpha-1)\cdot\sum_{cyc}x^{2}y^{2}+(\alpha+2)\cdot \sum_{cyc}x^{2}.$
If $ \alpha=7$ then for $ \sum_{cyc}\sqrt{\frac{a}{a+7b+c}}\geq1$ enough to prove that $ 27\geq12x^{2}y^{2}z^{2}+2(x^{2}y^{2}+x^{2}z^{2}+y^{2}z^{2})+3(x^{2}+y^{2}+z^{2}),$
where $ x+y+z=3,$ which equivalent to the following inequality:
$ \sum_{cyc}(x^{5}y+x^{5}z+x^{4}y^{2}+x^{4}z^{2}+9x^{4}yz+16x^{3}y^{2}z+16x^{3}z^{2}y-45x^{2}y^{2}z^{2})\geq0.$[/quote]

Yes, I have a quite same solution, but I just use the elementary method to construct the relationship between $ \alpha, \beta,\gamma$.

[quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers. Prove that

$ \sqrt {\frac {a}{b \plus{} 3c}} \plus{} \sqrt {\frac {b}{c \plus{} 3a}} \plus{} \sqrt {\frac {c}{a \plus{} 3b}}\ge\frac {3}{2}.$[/quote][quote="arqady"]Let $ \frac {a}{b \plus{} 3c} \equal{} \frac {x^{2}}{4},$ $ \frac {b}{c \plus{} 3a} \equal{} \frac {y^{2}}{4}$ and $ \frac {c}{a \plus{} 3b} \equal{} \frac {z^{2}}{4},$ where $ x,$ $ y$ and $ z$ are non-negative numbers.

Hence, it remains to prove that $ x \plus{} y \plus{} z\geq3,$ where $ 16 \equal{} 7x^{2}y^{2}z^{2} \plus{} 3(y^{2}z^{2} \plus{} z^{2}x^{2} \plus{} x^{2}y^{2}).$

Let $ x \plus{} y \plus{} z < 3.$ But it's contradiction since, [/quote]$ 7x^2y^2z^2 \plus{} 3(y^2z^2 \plus{} z^2x^2 \plus{} x^2y^2)$

$ < 7x^2y^2z^2\left(\frac {3}{x \plus{} y \plus{} z}\right)^6 \plus{} 3\left(y^2z^2 \plus{} z^2x^2 \plus{} x^2y^2\right)\left(\frac {3}{x \plus{} y \plus{} z}\right)^4$

$ \equal{} 16 \minus{} \sum{\frac {(y \minus{} z)^2}{(x \plus{} y \plus{} z)^6}\left[5x^4 \plus{} 732x^2yz \plus{} x(y \plus{} z)(5x^2 \plus{} 8y^2 \plus{} 253yz \plus{} 8z^2) \plus{} 8(y \plus{} z)^2(y^2 \plus{} 11yz \plus{} z^2)\right]}$

$ \leq 16.$

Other elementary solutions see : http://www.mathlinks.ro/viewtopic.php?t=105343
[quote="hungkhtn"]I think the following similar one is true, Arqady

$ \sqrt {\frac {a}{a \plus{} 2b \plus{} c}} \plus{} \sqrt {\frac {b}{b \plus{} 2c \plus{} a}} \plus{} \sqrt {\frac {c}{c \plus{} 2a \plus{} b}}\le \frac {3}{2}.$[/quote][quote="arqady"]Let $ \frac {a}{a \plus{} \alpha b \plus{} c} \equal{} \frac {x^{2}}{\alpha \plus{} 2},$ $ \frac {b}{b \plus{} \alpha c \plus{} a} \equal{} \frac {y^{2}}{\alpha \plus{} 2}$ and $ \frac {c}{c \plus{} \alpha a \plus{} b} \equal{} \frac {z^{2}}{\alpha \plus{} 2},$

where $ \alpha,$ $ x,$ $ y$ and $ z$ are non-negative numbers.

Then $ \begin{cases}(x^{2} \minus{} \alpha \minus{} 2)a \plus{} \alpha x^{2}b \plus{} x^{2}c \equal{} 0 \\
y^{2}a \plus{} (y^{2} \minus{} \alpha \minus{} 2)b \plus{} \alpha y^{2}c \equal{} 0 \\
\alpha z^{2}a \plus{} z^{2}b \plus{} (z^{2} \minus{} \alpha \minus{} 2)c \equal{} 0\end{cases}.$

Thus, $ \det\left( \begin{array}{ccc}x^{2} \minus{} \alpha \minus{} 2 & \alpha x^{2} & x^{2} \\
y^{2} & y^{2} \minus{} \alpha \minus{} 2 & \alpha y^{2} \\
\alpha z^{2} & z^{2} & z^{2} \minus{} \alpha \minus{} 2 \end{array}\right) \equal{} 0.$

Id est, $ (\alpha \minus{} 1)^{2}x^{2}y^{2}z^{2} \plus{} (\alpha \minus{} 1)\sum{y^{2}z^{2} } \plus{} (\alpha \plus{} 2)\sum{x^{2}} \equal{} (\alpha \plus{} 2)^{2} .$[/quote]If $ \alpha \equal{} 2$ then for $ \sum_{cyc}\sqrt {\frac {a}{a \plus{} 2b \plus{} c}}\leq\frac {3}{2}$ enough to prove that $ x^{2}y^{2}z^{2} \plus{} y^{2}z^{2} \plus{} z^{2}x^{2} \plus{} x^{2}y^{2} \plus{} 4(x^{2} \plus{} y^{2} \plus{} z^{2})\geq 16,$

where $ x \plus{} y \plus{} z \equal{} 3,$ which equivalent to the following inequality:

$ \sum{(y \minus{} z)^2\left[73x^4 \plus{} 85x^3(y \plus{} z) \plus{} 30x^2(y^2 \plus{} z^2) \plus{} 7x(y \plus{} z)(4y^2 \plus{} 11yz \plus{} 4z^2) \plus{} 10(y \plus{} z)^4\right]}$

$ \geq0.$

By the way, if $ a,b,c$ be positive numbers, then

$ \sqrt {\frac {a}{a \plus{} \alpha b \plus{} c}} \plus{} \sqrt {\frac {b}{b \plus{} \alpha c \plus{} a}} \plus{} \sqrt {\frac {c}{c \plus{} \alpha a \plus{} b}}\le \frac {3}{\sqrt {\alpha \plus{} 2}}$

holds if and only if $ 0\leq \alpha\leq\alpha_1 \equal{} 3.2953\cdots$, which the greatest root of $ 16\alpha^5 \minus{} 281\alpha^4 \plus{} 1825\alpha^3 \minus{} 2291\alpha^2 \minus{} 4130\alpha \plus{} 100,$

with equality if $ \alpha \equal{} \alpha_1,a \equal{} 1,b \equal{} 1.0853\cdots$ which the greatest root of $ b^5 \minus{} 3b^4 \plus{} 148b^3 \plus{} 677b^2 \minus{} 903b \minus{} 4,$

$ c \equal{} 1.3040\cdots$ which the greatest root of $ c^5 \plus{} 167c^4 \plus{} 7298c^3 \minus{} 16981c^2 \plus{} 8575c \plus{} 1024.$

$ \alpha \equal{} 3$ see : http://www.mathlinks.ro/viewtopic.php?t=299281

[quote="arqady"]
Probably, $ \sqrt {\frac {a}{a \plus{} 3b \plus{} c}} \plus{} \sqrt {\frac {b}{b \plus{} 3c \plus{} a}} \plus{} \sqrt {\frac {c}{c \plus{} 3a \plus{} b}}\le\frac {3}{\sqrt5}$ is true,
 but if it's true, it's not easy. :wink:[/quote]
Now it's easy enough!
By http://www.artofproblemsolving.com/Forum/viewtopic.php?p=893715#893715
for $ \alpha\equal{}3$ we need to prove that
\[ 25\leq4x^2y^2z^2\plus{}\sum_{cyc}(2x^2y^2\plus{}5x^2)\]
where $ x,$ $ y$ and $ z$ are non-negatives such that $ x\plus{}y\plus{}z\equal{}3.$
Let $ x\plus{}y\plus{}z\equal{}3u,$ $ xy\plus{}xz\plus{}yz\equal{}3v^2$ and $ xyz\equal{}w^3.$
Hence, $ 25\leq4x^2y^2z^2\plus{}\sum_{cyc}(2x^2y^2\plus{}5x^2)\Leftrightarrow f(w^3)\geq0,$ where
$ f(w^3)\equal{}2w^6\minus{}6u^3w^3\plus{}10u^6\minus{}15u^4v^2\plus{}9u^2v^4.$
But $ f'(w^3)\equal{}4w^3\minus{}6u^3\leq0.$
Hence, for the proof enough to check $ f(w^3)\geq0$ for  $ y\equal{}z\equal{}1,$
which gives $ (x\minus{}1)^2(10x^4\plus{}50x^3\plus{}87x^2\minus{}148x\plus{}244)\geq0,$ which is true.


Simplify $ f(n) \equal{}\sum_{k \equal{} 1}^{n}{2^{2^{k}\minus{}k}}$

$ f(n)\equal{}\int_{[0,2]}\sum_{k\equal{}1}^{n}x^{2^{k}\minus{}1}$ idon't know if it help.


inequalities
three variable inequality
algebra
IMO Shortlist

I think so. The solution without typos :

Note that $a^5 + b^5 \geq a^3b^2 + a^2b^3$ by Muirhead, rearrangement, AM - GM, ... so

\begin{eqnarray*} \sum \frac{ab}{a^5 + b^5 + ab} &\leq& \sum \frac{ab}{a^3b^2 + a^2b^3 + ab} \\ \ &=& \sum \frac{1}{a^2b + ab^2 + 1} \\ \ &=& \sum \frac{abc}{a^2b + ab^2 + abc } \\ \ &=& \sum \frac{c}{a + b + c} \\ \ &=& 1. \end{eqnarray*}

We're done ! :cool:

One can easily show that $ x^5 \plus{} y^5 \geq x^2 y^3 \plus{} x^3 y^2$ for every x, y positive

$ ab/(a^5\plus{}b^5\plus{}ab) \leq ab/(a^2 b^3 \plus{}a^3 b^2\plus{}ab) \equal{} 1/(a b^2\plus{}a^2 b\plus{}1) \equal{} 1/[ab(a\plus{}b\plus{}c)]  \equal{}c/(a\plus{}b\plus{}c)$ since $ abc \equal{}1$

Similarly  $ bc/(b^5\plus{}c^5\plus{}bc) \leq a/(a\plus{}b\plus{}c)$ and  $ ca/(c^5\plus{}a^5\plus{}ca)\leq b/(a\plus{}b\plus{}c)$

By adding those three inequalities we get the result we are looking for.

Cute no?

$ \frac {xy}{x^5 \plus{} y^5 \plus{} xy} \equal{} \frac {xyz}{x^5z \plus{} y^5z \plus{} xyz} \equal{} \frac {1}{x^5z \plus{} y^5z \plus{} 1}$
Now, $ \frac {1}{a\plus{}1} \plus{} \frac {1}{b\plus{}1} \plus{} \frac {1}{c\plus{}1} \implies a \plus{} b \plus{} c \plus{} 2 \le abc$
Thus, we have to show
$ 2 \plus{} \sum {x^5y} \equal{} x^5y \plus{} x^5z \plus{} y^5x \plus{} y^5z \plus{} z^5x \plus{} z^5y \plus{} 2 \le (x^5 \plus{} y^5)(y^5 \plus{} z^5)(x^5 \plus{} z^5) \equal{} 2 \plus{} \sum {x^{10}y^5}$
But $ \sum{x^5y} \equal{} \sum{x^8y^4z^3}$, and we have Muirhead.

1998,I proof:
Suppose that $a, b, c > 0$ such that $abc = 1$ ,  $n$ be positive integer. Prove that
\[ \frac{a^{n-1}b^{n-1}}{a^{n-1}b^{n-1} + a^{2n+1} + b^{2n+1}} + \frac{b^{n-1}c^{n-1}}{b^{n-1}c^{n-1} + b^{2n+1} + c^{2n+1}}
\[\\ + \frac{c^{n-1}a^{n-1}}{c^{n-1}a^{n-1} + c^{2n+1} + a^{2n+1}} \leq 1. \]
Another generalization ：
Let  $a,b,c $ be positive real numbers such that $abc=1 ,n\le-1 $   or  $ n\ge2 ,$ Show that\[\frac{bc}{b^n+c^n+bc}+\frac{ca}{c^n+a^n+ca}+\frac{ab}{a^n+b^n+ab}\le1\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2835835

Similarly have
Let  $a,b,c>0 $ such that $abc=1 ,n\ge1 $   or  $ n\le-2 ,$ Show that
\[\frac{a^n}{a^n+b+c}+\frac{b^n}{b^n+c+a}+\frac{c^n}{c^n+a+b}\ge1\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&p=2724850#p2724850

Notice that we can rewrite given $LHS$ as \[\sum_{cyc}{\frac{ab}{ab+a^5+b^5}}=\sum_{cyc}{\frac{1}{1+a^5c+b^5c}}.\]
Then using supstiution $a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}$ we get \[\sum_{cyc}{\frac{1}{1+a^5c+b^5c}}=\sum_{cyc}\frac{1}{1+\frac{x^4z}{y^5}+\frac{y^5}{z^4x}}\leq \sum_{cyc}\frac{1}{2\cdot \sqrt\frac{x^3}{z^3}+1}\] which we got by $AM-GM$.
Then we have \[\sum_{cyc}\frac{1}{2\cdot \sqrt\frac{x^3}{z^3}+1}\leq\sum_{cyc}\frac{1}{2\cdot \sqrt\frac{x^3}{x^3}+1}=\sum_{cyc}\frac{1}{2+1}=\frac{1}{3}\cdot 3=1.\]Thats true because of rearrangement.
$Q.E.D.$

@above: I don't see how you applied rearrangment in the final line. Also many solutions here use $a^5+b^5 \ge a^3b^2+b^2a^3$ (*). I have seen some times before the trick where you reduce an inequality to cylcic sum of $a/(a+b+c)$ equals 1, and if you have this trick in mind then it is quite easy to find that you need to prove (*), and then the problem is solved. But it seems to me a very unnatural thing to think of the trick first, and to me its not obvious at all that the trick is applicable here at first, so it does not come to mind. My question is then: how did you think to use this inequality (*)?

@above: your post has nothing do with the problem in this thread. you should remove it i think.

[quote=velraman][quote=Mewto55555]After homogenizing by turning the $ab$ terms into $a^2b^2c$ and etc and multiplying by 2, straight-up expansion gives the LHS to be

$[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$

and the RHS to be

$[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$

Subtracting gives $RHS-LHS=2[10,5,0]-2[8,4,3]$ but $[10,5,0] \geq [8,4,3]$ by Muirhead so we're done.[/quote]

I don't know what he mean by $[11,2,2]+4[8,4,3]+[7,7,1]+2[7,6,2]+[5,5,5]$ and $[11,2,2]+2[10,5,0]+[7,7,1]+2[7,6,2]+2[8,4,3]+[5,5,5]$

Can someone explain ?[/quote]

He is probably referring to the coefficients, and the ordered triplet represents the powers of a,b, and c respectively.

My Solution :

Observe that $(x^3-y^3)(x^2-y^2) \ge 0 \implies x^5 + y^5 \ge x^2y^2(x+y) \ \forall x,y \in \mathbb{R^+}$.

Thus, $\sum_{cyc} \dfrac {ab}{a^5+b^5+ab} \le \sum_{cyc} \dfrac{ab}{a^2b^2(a+b)+ab} = \sum_{cyc} \dfrac{c}{a+b+c} = 1$.

\[(ab+a^5+b^5)\left(\frac{c^4}{ab} + \frac1a + \frac1b\right)\ge (a^2+b^2 + c^2)^2\]
\[\iff \frac{ab}{ab+a^5+b^5} \le \frac{c^4 + a + b}{(a^2 + b^2 + c^2)^2}\]
Summing up similar inequalities.
\[\text{LHS} \le \frac{\sum a^4 +2\sum a }{(a^2 + b^2 + c^2)^2}\]
So, it suffices to prove that
\begin{align*}
\frac{\sum a^4 +2\sum a }{(a^2 + b^2 + c^2)^2}&\le 1\\
\iff a^2b^2 + b^2c^2 + c^2a^2 &\ge a + b + c\\
&= abc(a+b+c)\\
\iff \sum a^2(b-c)^2 &\ge 0
\end{align*}

[quote=Arne]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that  \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \][/quote]

$$\sum\frac{ab}{ab+a^5+b^5}=\sum\frac{a^2b^2c}{a^2b^2c+a^5+b^5}\leq\sum\frac{a^2b^2c}{a^2b^2c+a^3b^2+a^2b^3}=\sum\frac{c}{a+b+c}=1$$ :)

Observe that (By Muirhead) $a^5+b^5\ge a^2b^2(a+b)$. We have
\begin{align*}
\sum_{cyc}\frac{ab}{a^5+ab+b^5}&\le\sum_{cyc}\frac{ab}{a^2b^2(a+b)+ab} \\
&=\sum_{cyc}\frac1{ab(a+b)+1} \\
&=\sum_{cyc}\frac1{ab(a+b)+abc} \\
&=\sum_{cyc}\frac{1}{ab(a+b+c)} \\
&=\sum_{cyc}\frac c{a+b+c} \\
&=1
\end{align*}

Very similar to a problem from USAMO 1998.

[quote=Arne]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that  \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \][/quote]
Let $a,b,c>0$ with $abc=1.$  [url=https://artofproblemsolving.com/community/c6h2071085p14829277]Establish that[/url]
$$\frac{1}{a^5+b^5+1}+\frac{1}{b^5+c^5+1}+\frac{1}{c^5+a^5+1}\leq 1.$$
$$\iff$$
$$\frac{1}{a^3+b^3+1} +\frac{1}{b^3+c^3+1} +\frac{1}{c^3+a^3+1} \leq  1$$
[url=https://artofproblemsolving.com/community/c6h578290p3411216]h[/url] [url=https://artofproblemsolving.com/community/c6h1305663p6976340]h[/url] [url=https://artofproblemsolving.com/community/c4h1883433p12823952]h[/url]

By Muirhead’s inequality, $b^5 + c^5 \ge b^3c^2 + b^2c^3$. Hence,
$$\sum_{cyc} \frac{ab}{ab+a^5+b^5} \le \sum_{cyc} \frac{ab}{ab+a^3b^2 + a^2b^3} = \sum_{cyc} \frac{1}{1+a^2b+b^2a} = \sum_{cyc} \frac{c}{c+a+b} = 1$$
as desired.

WLOG let $a$$\leq$$ b$$ \leq c$. So we have, from rearrangement inequality, $\begin{bmatrix} a^3 & b^3 \\ a^2 & b^2 \end{bmatrix} \geq \begin{bmatrix} a^3 & b^3 \\ b^2 & a^3 \end{bmatrix}$. Which in turn gives , $a^5+b^5 \geq a^{3}b^{2} + a^{2}b^{3} $ \\ $\frac {ab}{a^5+b^5 +ab} $$\leq$$ \frac {ab}{ a^{3}b^{2} + a^{2}b^{3}+ab}$\\ \\ Then it follows :)

Note that
$\frac{xy}{x^5+ xy+y^5}=\frac{1}{z(x^5+y^5)+1} \leq \frac{1}{z(x^4y+xy^4)+1}=\frac{1}{(x^3+y^3+1)} \leq \frac{1}{xy(x+y)+1}=\frac{z}{x+y+z}$ and the result follows by summing this cyclically $\Box$

$a^5+b^5\ge a^4b+ab^4$ and $a^3+b^3\ge a^2b+ab^2$ are direct by Rearrangement, so we have:
$\sum_{\text{cyc}}\frac{ab}{ab+a^5+b^5}\le\sum_{\text{cyc}}\frac1{1+a^3+b^3}\le\sum_{\text{cyc}}\frac c{c+a^2bc+ab^2c}=\sum_{\text{cyc}}\frac a{a+b+c}=1$

The classic Muirhead for denominator inequalities!

We have that $\sum_{\text{cyc}} \frac{ab}{ab+a^5+b^5}\leq \sum_{\text{cyc}} \frac{ab}{ab+a^3b^2+a^2b^3} = \sum_{\text{cyc}} \frac{1}{abc+a^2b+ab^2}=\frac{a+b+c}{a+b+c}=1.$

Remarks: In the beginning, I tried to use AM-GM directly on the denominator to get that I wanted to show that $\sum_{cyc} \frac{1}{ab}\leq 3\implies a+b+c\leq 3,$ but obviously the inequality was too weak. :(

\begin{align*}
\sum_{\text{cyc}} \dfrac{ab}{{\color{red}a^5+b^5}+ab} &\le \sum_{\text{cyc}} \dfrac{ab}{{\color{blue}a^3b^2 + a^2b^3} + ab} \\
&= \sum_{\text{cyc}} \frac{ab}{ab(a^2b + ab^2 + {\color{red}1})}\\
&= \sum_{\text{cyc}} \frac{1}{a^2b + ab^2 + {\color{blue}abc}} \\
&= \sum_{\text{cyc}} \frac{1}{{\color{red}ab}(a+b+c)} \\
&= \sum_{\text{cyc}} \frac{\color{blue}c}{a+b+c}\\
&= 1
\end{align*}

By Muirhead's Theorem: $a^5+b^5 \geq a^2b^3+a^3b^2$
$\sum_{\text{cyc}} \frac{ab}{a^5+b^5+ab}\leq \sum_{\text{cyc}} \frac{ab}{a^3b^2+a^2b^3+ab} = \sum_{\text{cyc}} \frac{1}{a^2b+ab^2+1}=\sum_{\text{cyc}}\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1$.$\blacksquare$

[quote=Arne]Suppose that $a, b, c > 0$ such that $abc = 1$. Prove that  \[ \frac{ab}{ab + a^5 + b^5} + \frac{bc}{bc + b^5 + c^5} + \frac{ca}{ca + c^5 + a^5} \leq 1. \][/quote]

$(a^3-b^3)(a^2-b^2) \geq 0 \implies a^5+b^5 \geq a^2b^2(a+b),$ equality when $a=b$. So, $\frac{ab}{a^5+b^5+ab} \leq \frac{ab}{a^2b^2(a+b)+ab} = \frac{1}{ab(a+b)+1} =\frac{abc}{ab(a+b+c)} =\frac{c}{a+b+c}.$
Similarly, $\frac{bc}{b^5+c^5+bc} \leq \frac{a}{a+b+c}$ and $\frac{ca}{c^5+a^5+ca} \leq \frac{b}{a+b+c},$ equality when $a=b=c=1.$
The claim follows adding these three. $\square$

Same as above solutions; posting for storage.

By Muirhead, $a^5+b^5\ge a^3b^2+b^3a^2.$ Hence, $$\sum_{\text{cyc}}\frac{ab}{ab+a^5+b^5}\le\sum_{\text{cyc}}\frac{ab}{ab+a^3b^2+b^3a^2}=\sum_{\text{cyc}}\frac{1}{\frac{a}{c}+\frac{b}{c}+\frac{c}{c}}=\sum_{\text{cyc}}\frac{c}{a+b+c}=1.$$ $\square$

Every solution is going to look the same I guess.
We first show that $a^5+b^5 \ge a^3b^2++b^3a^2$. This will be just expanding and applying AM-GM inequality.
\begin{align*}
a^5+b^5 = (a+b)(a^4-a^3b+a^2b^2-ab^3+b^4) \ge (a+b)(2a^2b^2+a^2b^2-ab(a^2+b^2)) \ge (a+b)(3a^2b^2-2a^2b^2) = a^3b^2+b^3a^2
\end{align*}
Returning to the problem:
\begin{align*}
\sum_{\text{cyc}}\frac{ab}{ab+a^5+b^5} = \sum_{\text{cyc}}\frac{1}{1+c(a^5+b^5)} \le \sum_{\text{cyc}}\frac{1}{1+c \cdot a^2b^2(a+b)} = \sum_{\text{cyc}}\frac{1}{1+ab(a+b)} = \sum_{\text{cyc}}\frac{c}{a+b+c} =1
\end{align*}
which means we are done. $\blacksquare$

[quote=jasperE3]For positive reals $x,y,z$ with $xyz=1$, prove or disprove that:
$$\frac1{x+y+1}+\frac1{y+z+1}+\frac1{z+x+1}\le1.$$[/quote]

Interesting. You can easily prove this result is true from straight-up expansion and then Muirhead.

Note that this is actually equivalent to the original problem. From the substitution $x=a^5b$, $y=b^5c$, etc, the LHS turns into $\sum_{cyc}{\frac{1}{ab^5 + bc^5 + 1}}$, and dividing both top and bottom by $abc=1$ gives the original inequality.

------

I would be curious to see if anyone has a solution from Holder's/Cauchy's/Titu's (from $\sum_{cyc}{\frac{a^5+b^5}{ab+a^5+b^5}} \ge 2$.) I tried but didn't get anywhere.

By Power-Mean Inequality, we obtain
$$\sqrt[5]{\frac{a^5+b^5}{2}} \geq \frac{a+b}{2}$$
\begin{align*}
\implies \frac{a^5+b^5}{2} &\geq \left(\frac{a+b}{2}\right)^5 \\
&\geq \left(\frac{a+b}{2}\right)^4 \cdot \frac{a+b}{2} \\
& \geq a^2b^2\left(\frac{a+b}{2}\right) \text{[using AM-GM]}\\
\end{align*}
$$\therefore \boxed{a^5+b^5\geq a^2b^2(a+b)}.$$

Hence,
\begin{align*}
\sum_{cyc} {\frac{ab}{a^5+b^5+ab}} &\leq \sum_{cyc} {\frac{ab}{a^2b^2(a+b)+ab}} \\
&= \sum_{cyc} {\frac{1}{ab(a+b)+1}} \\
&= \sum_{cyc} {\frac{c}{a+b+c}} \text{[using $abc=1$]} \\
&= 1.
\end{align*}
$QED.$

Solved with [b]CT17[/b].

Notice that $(b^3-c^3)(b^2-c^2) \ge 0 \implies b^5+c^5 \ge b^3c^2+b^2c^3$, so we have \[\sum_\text{cyc} \frac{bc}{bc+b^5+c^5} \le \sum_\text{cyc} \frac{bc}{bc+b^3c^2+b^2c^3}=\sum_\text{cyc} \frac{\frac{1}{a}}{\frac{1}{a}+\frac{b}{a^2}+\frac{c}{a^2}}=\sum_\text{cyc} \frac{a}{a+b+c}=1,\] as desired. $\square$

[b]Version 1[/b]
Let $a,b,c$ be positive reals such that $abc=2$. Then prove that

$$\sum_{cyc}{\dfrac{a^2b^2}{2\left(a^2b^2\right)+a^{7}+b^{7}}}\leq \dfrac{1}{2}$$

[b]Generalization 1[/b]
Let $a,b,c$ be positive reals such that $abc=k$. Then prove that

$$\sum_{cyc}{\dfrac{(ab)^{n}}{k\left(ab\right)^{n}+a^{2n+3}+b^{2n+3}}}\leq \dfrac{1}{k}$$

From rearrangement inequality, we have: $a^5+b^5\geq a^3b^2+a^2b^3 \implies a^5+b^5\geq a^2b^2(a+b)$
$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{ab}{a^2b^2(a+b)+ab}}$$
$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{1}{ab(a+b)+1}}$$
$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{abc}{ab(a+b)+abc}}$$
$$\implies \sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq \sum_{cyc} {\frac{c}{a+b+c}}$$
$$\implies \boxed{\sum_{cyc} {\frac{ab}{a^5+b^5+ab}} \leq 1}$$


geometry
geometric transformation
reflection
integration
real analysis
real analysis unsolved

Here is a beautiful problem.  I already have a solution, but I'm looking for other (creative) solutions.

[i]Find the area of the region
\[R=\{(x,y)\in\mathbb{R}^{2}: \sqrt{x}+\sqrt{y}\geq 1,\ \sqrt{1-x}+\sqrt{1-y}\geq 1\}. \][/i]
Masoud Zargar

Here's something straightforward. The set in question is contained in the square $[0,1]\times[0,1].$ The transformation $(x,y)\mapsto(1-y,1-x)$ reflects the region about the line $x+y=1$ but leaves it unchanged. The region lies between two curves. What I will do is calculate the area below the lower curve:

$\int_{0}^{1}\left(1-\sqrt{x}\right)^{2}\,dx=\int_{0}^{1}1-2\sqrt{x}+x\,dx=\frac16.$

The region above the upper curve has the same area. We can find the area we want by subtraction:

Area = $1-\frac16-\frac16=\frac23.$


[quote="Beat"]Is it true that positive reals sattisfy
$\sqrt{\frac{(x+y+z)^3}{x^2+y^2+z^2+xy+yz+zx}}\ge \frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{2}}$[/quote]

Alas, it isn't: try x = 1, y = 1, z = 3.

  Darij


algebra
polynomial
rotation
analytic geometry
function

I was asked to post these once they were available, so here they are. The solutions are online as well, but have a bit of restraint and try to solve the problems rather than just searching for solutions.

1. Consider the equation
x1x2 + x2x3 + x3x4 +    + xn−1xn + xnx1 = 0
where xi is selected from the set {1,−1} for i = 1, 2, . . . , n.
(a) Show that if the equation has a solution, then n is even. (2 points)
(b) Suppose n is divisible by 4. Show that the equation has a solution. (2 points)
(c) Show that if the equation has a solution, then n is divisible by 4. (6 points)

2. (a) Find a polynomial f(x) with integer coeffcients and two distinct integers a and b such that
f(a) = b and f(b) = a. (2 points)
(b) Let f(x) be a polynomial with integer coeffcients and a, b, and c be three integers. Suppose f(a) = b, f(b) = c, and f(c) = a. Show that a = b = c. (8 points)

3. (a) Consider the triangle with vertices M(0, 2n + 1), S(1, 0), and U(0, 1/(2n :^2:)), where n is a positive
integer. If :theta: = 6 MSU, prove that tan :theta: = 2n − 1. (5 points)
(b) Find positive integers a and b that satisfy the following equation. (2 points)
arctan 1/8 = arctan a − arctan b
(c) Determine the exact value of the following infinite sum. (3 points)
arctan 1/2 + arctan 1/8 + arctan 1/18 + arctan 1/32+    + arctan 1/(2n :^2:) +

4. (a) Prove: (55 + 12*:sqrt:21)1/3 + (55 − 12*:sqrt:21)1/3 = 5. (5 points)
(b) Completely factor x8 + x6 + x4 + x2 + 1 into polynomials with integer coefficients, and explain
why your factorization is complete. (5 points)

5. In this problem, we simulate a hula hoop as it gyrates about your waist. We model this situation by
representing the hoop with a rotating a circle of radius 2 initially centered at (−1, 0), and representing
your waist with a fixed circle of radius 1 centered at the origin. Suppose we mark the point on
the hoop that initially touches the fixed circle with a black dot (see the left figure).
As the circle of radius 2 rotates, this dot will trace out a curve in the plane (see the right figure).
Let :theta: be the angle between the positive x-axis and the ray that starts at the origin and goes through
the point where the fixed circle and circle of radius 2 touch. Determine formulas for the coordinates
of the position of the dot, as functions x(:theta:) and y(:theta:). The left figure shows the situation when :theta: = 0
and the right figure shows the situation when :theta: = 2 :pi: /3. (10 points)  (Sorry, guys, I can't attach the diagrams easily, so visit http://www.delta.edu/math/mmpc/graphics/exams/2003.II.problems.pdf if you need them or if there are typos in any of the previous problems)

1. [hide]a. Clearly, half the terms will have to be -1 and half the terms 1. As there are n terms, n is even. b. let those with indices congruent to 0 or 1 mod 4 be 1, and those with indices congruent to 2 or 3 mod 4 be -1. Solution follows. c. consider changing one of the "-1"s from -1 to 1. It will change the sum by either -4, 0, or 4, so the sum remains constant mod 4. As the sum was 0 mod 4 at the beginning, when all the terms are 1, we have n:equiv:0 mod 4.[/hide]

2. [hide]a. f(x)=1-x. try x=0, 1 b. If f(a)=b and f(c)=d and f is a polynomial function, (c-a)|(d-b). In this case, we have (b-a)|(c-b), (c-b)|(a-c), (a-c)|(b-a). We thus have (b-a)=:pm:(c-b), and cyclic permutations of that equation. That equations simplifies to a=c OR a+c=2b. Anyway, it's easy enough from here (remember we also have cyclic permutations of those equations) to come up with a=b=c.[/hide]

4. [hide]a. let x=that thing. x^3=55+12:sqrt:21+55-12:sqrt:21+3x[(55+12:sqrt:21)(55-12:sqrt:21)]^(1/3)=110+3x => x^3-3x-110=0. This factors into (x-5)(x^2+5x+22)=0. This has roots x=5, and some complex roots. However, that thing was obviously real. Thus x=5. b. Using fifth roots of unity, we find that if we plug in any fifth root of unity, we get 0. Thus, (x^4+x^3+x^2+x+1) must factor that. Indeed it does, and it factors into (x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1). Now, it is easy to see that this is a complete factorization because that can be written as (x^5-1)(x^5+1)/(x-1)(x+1), and the only things that divide that are complex roots of -1 and 1. [/hide]

3. [hide]a. in the problem statement, :theta:=MSU, not 6MSU, so i'll assume the problem is right and that was just a typo. in any case, (O is the origin in case you didn't guess) :theta:=OSM-OSU, and tan:theta:=[(2n+1)-(1/2n^2)]/[1+(2n+1)(1/2n^2)]=[4n^3+2n^2-1]/[2n^2+2n+1]=2n-1. b. 8=(1+ab)/(a-b) => 8a-8b=1+ab => ab-8a+8b=-1 => (a+8)(b-8)=-65. (a,b)=(57, 7) is one solution methinks. It's easy to find all solutions but I won't. c. by (a), arctan(1/2n^2)=arctan(2n+1)-arctan(2n-1) so this whole thing is arctan(:inf:) (you know that means limit of arctan n as n goes to infinity) - arctan (1)= :pi:/2-:pi:/4=:pi:/4[/hide]

5. [hide]at :theta:, the circles intersect (are tangent, whatever) at (cos :theta:, sin :theta:). The center of the big circle will be the negative of that. Okay, now if the big circle were centered at the origin, the marked point would be at (2cos :theta:/2, 2sin :theta:/2). We now translate that by the negative of that original point/vector to get (x(:theta:), y(:theta:))=(2cos:theta:/2-cos:theta:, 2sin:theta:/2-sin:theta:)[/hide]

does everyone hate my solutions so much they don't even want to discuss them...

Mystic, I told you they looked good... and got some information out of you on AIM ;) What should I say here?

oh sorry i forgot. i was hoping people might possibly have other solutions though but whatever.


When the temperature goes up $ 3^\circ$ on the Cantor scale, it goes up $ 8^\circ$ on the Frobenius scale. On both scales, $ 18^\circ$ is the same temperature. How many Frobenius degrees are equal to $ 30^\circ$ Cantor?

When the temperature goes up  3on the Cantor scale, it goes up  8 on the Frobenius scale.
When it goes up 12 on Cantor (18 to 30), it goes up 32 on the Frobenius scale.
18+32=[b]50[/b]


function
trigonometry
limit
real analysis
real analysis unsolved

Show that continuity of the differential is essential in the inverse function theorem by considering the function $ f: (\minus{}1,1)\to\mathbb{R}$ defined by $ f(0)\equal{}0$ and $ f(t)\equal{}t\plus{}2t^2\sin(1/t)$ for $ t\neq0$. Show that $ f$ is everywhere differentiable, and even that $ f'$ is bounded, that $ f'(0)\equal{}1$, but that $ f$ is not injective in any neighborhood of $ 0$

Which part are you having trouble with? For example:
$ f'(t) \equal{} 1 \plus{} 4t\sin\left(\frac {1}{t}\right) \minus{} 2\cos\left(\frac {1}{t}\right)$ for $ t\neq 0$. Since  $ |t| < 1$ and sine and cosine are bounded anyway, $ |f'(t)|$ is bounded.
And:
$ f'(0) \equal{} \lim_{x\to 0}\frac {f(x) \minus{} f(0)}{x} \equal{} \lim_{x\to 0}\frac {x \plus{} 2x^2\sin\left(\frac {1}{x}\right)}{x} \equal{} \lim_{x\to 0}1 \plus{} 2x\sin\left(\frac {1}{x}\right) \equal{} 1$. Ehm..I'm doing something wrong here apparently (it should be $ 0$).


geometry
geometric transformation
rotation

examin the relationship below. Determine whcih of the following properties each relationship satisfies: the Reflexive Property (a is a real number, a=a), the symmetric property(a=b, b=a). Transitive property(a=b, b=c, a=c). Then state wether the relationship is an equivalence relation. If a preperty is not satisfied by a given relationship, give a counterexample.

1 A less than B, where A and B are real numbers

2.A less or equal than B( <= ), A and B are real numbers

3.A is disvisible by b, where a and b are real numbers

4.Figure a is a relfection of figure b.

5.Firgure a is a rotation of firgure b

6. A is a sister of B, where a and b are people.

53. A has the same last name as b, where a and b are people

[hide]
1. Transitive, not equivalence
2. Equivalence (in some cases)
3. Transitive, reflexive, not equivalence
4. symmetric, not equivalence
5. Reflexive, symmetric, not equivalence
6. Transitive, not equivalence
53. Equivalence
[/hide]

Based on the content of this problem, I think it should go in Puzzles and Brainteasers. I'm not a moderator, though, so I can't move it.


Give a,b,c is three side of a triangle and $ a\plus{}b\plus{}c\equal{}2$.Prove that:
$ a^2\plus{}b^2\plus{}c^2\plus{}2abc<2$

Setting $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$ and after homogenization the inequality becomes:

$ 2 \sum x^3 \plus{} 6 \sum _{sym}x^2y \plus{} 10 xyz < 2\sum x^3 \plus{} 6 \sum _{sym} x^2y \plus{} 12xyz \Longleftrightarrow 2xyz > 0$ which is true  :wink:

Another solution:

By triangle ineq. we have $ a<b\plus{}c$ or $ 2a<a\plus{}b\plus{}c\equal{}2$, so $ a<1$ . Similarly obtain $ b<1$ and $ c<1$ .

Now it's obvious that $ (a\minus{}1)(b\minus{}1)(c\minus{}1)<0\Longleftrightarrow abc\minus{}ab\minus{}bc\minus{}ca\plus{}a\plus{}b\plus{}c\minus{}1<0$

$ \Longleftrightarrow abc\plus{}1<ab\plus{}bc\plus{}ca \Longleftrightarrow 2abc\plus{}2<2(ab\plus{}bc\plus{}ca)$

Adding left-right $ a^2\plus{}b^2\plus{}c^2$ it follows that $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\plus{}2<(a\plus{}b\plus{}c)^2$

$ \Longleftrightarrow a^2\plus{}b^2\plus{}c^2\plus{}2abc<2$, so we are done .


geometry
modular arithmetic
geometric transformation
reflection
geometry unsolved

$ O$ is a point in triangle $ ABC$.Let $ P,Q,R$ be the feets of the perpendiculars from $ O$ to $ BC,CA,AB$,respectively. $ A_1,B_1,C_1$ be three points in $ BC,CA,AB$ and $ A_2,B_2,C_2$ be the symetric of $ A_1,B_1,C_1$ by $ P,Q,R$.
$ Z_1,Z_2$ be the $ Miquel$ points of triangle $ ABC$ with $ A_1,B_1,C_1$ and $ A_2,B_2,C_2$. Prove that
$ OZ_1 \equal{} OZ_2$.

Let $ X\equal{}A_1Z_1\cap OP, Y\equal{}OQ\cap B_1Z_1, Z\equal{}OR\cap C_1Z_1$
It's easy to see that $ \angle Z_1ZO\equal{} \angle Z_1XO\equal{} \angle Z_1YO$ then $ Z_1,O,X,Y,Z$ lie on a circle.
Let $ XA_2\cap YB_2\equal{}\{Z'_2\}$
We have $ \angle B_2YQ\equal{}\angle B_1YQ\equal{}\angle A_1XP\equal{}\angle Z_2XP \Rightarrow Z'_2$ lies on $ (OXY)$
$ \Rightarrow \angle Z'_2YO\equal{}\angle Z'_2ZO\equal{}\angle OXA_2\equal{}\angle RZC_1\equal{}\angle RZC_2$
$ \Rightarrow C_2, Z, Z'_2$ are collinear.
$ \Rightarrow \angle Z'_2C_2A\equal{}\angle Z'_2B_2C\equal{}\angle Z'_2Z_2B \Rightarrow Z'_2$ is the Miquel point of triangle $ ABC$ wrt $ (A_2, B_2, C_2)$
$ \Rightarrow Z'_2\equiv Z_2$
But $ \angle Z_2ZO\equal{}\angle Z_1ZO$ so $ OZ_1\equal{}OZ_2$

[quote="thanhtra_dhsp"][b]Elementaryyy[/b] stealing this problem of [b]thuylinhk41[/b][/quote]
Oh don't say that! [b]Elementaryyy[/b] didn't tell this problem was his own. Moreover, as a file [b]plane geometry[/b] uploaded, this problem is not your.

[quote="thanhtra_dhsp"][b]Elementaryyy[/b] stealing this problem of [b]thuylinhk41[/b][/quote]
I think it is not your.

Another solotion:
We needs lemma
Lemma. Let two circle $ (O, R_a), (O, R_b)$. The points $ A_1, B_2\in(O, R_a);\ B_1, B_2\in (O, R_b)$. Call $ K\equiv (A_1A_2\cap B_1B_2)$; $ E_1\in (KA_1B_1)$ and $ KE\cap (O, OE)\equiv E_2$. We have $ E\in (KA_2B_2)$.

Return the problem
Call $ J$ is interssection point of $ AA_3$ and $ BB_3$
We have:

$ (FA_3, FB_3)$
$ \equiv (FA_3, FC_2) \plus{} (FC_2, FB_3)\pmod\pi$
$ \equiv (B_2A_3, B_2C_2) \plus{} (A_2C_2, A_2B_3)\pmod\pi$
$ \equiv (AA_3, AC_2) \plus{} (BC_2,BB_3)\pmod\pi$
$ \equiv (AA_3, BB_3)\pmod\pi$
$ \equiv(JA_3, JB_3)$

Thus $ J\in (FA_3B_3C_3)$.
Easily prove $ AA_3, BB_3, CC_3$ concurrent $ J$.
By the lemma $ J$ in $ (AB_2C_2), (BC_2A_2), (CA_2B_2)$.
Because $ E \equiv (AB_1C_1)\cap (BC_1A_1)\cap (CA_1B_1)$, so $ E\equiv J$.
Sumary $ OE \equal{} OF$.

Hint to another solution :
Reflect point A_1 wrt perpendicular thru point O to A_1Z_1 and get point A_2'
Like the same construct points B_2' , C_2'
Not hard to prove that A_2'B_2'C_2'O =  A_2B_2C_2O


Let   $ f: M_{nxn}(R) \to M_{nxn}(R)$ be defined by $ f(A) \equal{} 2009A \plus{} A^{T}$
1)Prove f is the linear transformation.
2)Find $ \det(f)$
Here $ A^{T}$ be transpose of A.

1) We will prove that $ f(A\plus{}\alpha B)\equal{}f(A)\plus{}\alpha f(B)$ for all $ \alpha\in\mathbb{R}$. $ f(A\plus{}\alpha B)$ equals $ 2009(A\plus{}\alpha B)\plus{}(A\plus{}\alpha B)^\tau$. Since $ (X\plus{}Y)^\tau\equal{}X^\tau\plus{}Y^\tau$ and we can use distribution property, this is equal to $ 2009A\plus{}A^\tau\plus{}\alpha(2009B\plus{}B^\tau)$, which, of course, equals $ f(A)\plus{}\alpha f(B)$.

3) Prove that $ f$  is diagonalizable and find eigenvalues of $ f$.
 :wow:

[quote="QuyBac"]Let   $ f: M_{nxn}(R) \to M_{nxn}(R)$ be defined by $ f(A) \equal{} 2009A \plus{} A^{T}$
1)Prove f is the linear transformation.
2)Find $ \det(f)$
3) Prove that f is diagonalizable and find eigenvalues of f .
Here $ A^{T}$ be transpose of A.[/quote]
 You can use $ M_{n}(R)\equal{}S_n(R) \bigotimes A_n(R)$
     $ S_n(R)$ are set real symmetric matrix.
     $ A_n(R)$ are set real skew-symmetric matrix.
 Easy : {$ (E_{ij}\plus{}E_{ji})_{1 \leq i<j \leq n}$} $ \bigcup$$ {(E_{ii})_{1 \leq i \leq n}}$ is basis of $ S_n(R)$
           {$ (E_{ij}\minus{}E_{ji})_{1 \leq i<j \leq n}$} is basis of $ A_{n}(R)$
Put $ \alpha \equal{} 2009$ when : $ f(A)\equal{}(\alpha\minus{}1)A\plus{}(A\plus{}A^T)\equal{}\frac{\alpha\plus{}1}{2}(A\plus{}A^T)\plus{}\frac{\alpha\minus{}1}{2}(A\minus{}A^T)$
 Implise :$ f((E_{ij}\plus{}E_{ji})_{1 \leq i \leq j \leq n})\equal{}(\alpha\plus{}1)(E_{ij}\plus{}E_{ji})_{1 \leq i \leq j \leq n}$
              $ f((E_{ij}\minus{}E_{ji})_{1 \leq i<j \leq n})\equal{}(\alpha\minus{}1)(E_{ij}\minus{}E_{ji})_{1 \leq i<j \leq n}$
Find matrix of $ f$ from basis of $ M_{n}(R)$
         Implies 3)2) $ \det(f)\equal{}(\alpha\plus{}1)^{dim(S_n(R))}(\alpha\minus{}1)^{dim(A_n(R))}\equal{}(\alpha\plus{}1)^{\frac{n(n\plus{}1)}{2}}(\alpha\minus{}1)^{\frac{n(n\minus{}1)}{2}}$ !
------------------------
hi... [b] ILY [/b]  :bruce:


The sides of a right triangle are a and b and the hypotenuse is c. A perpendicular from the vertex divides c into segments r and s, adjacent respectively to a and b. If a : b = 1 : 3, then the ratio of r to s is:

\textbf{(A)}\ 1 : 3\qquad \textbf{(B)}\ 1 : 9\qquad \textbf{(C)}\ 1 : 10\qquad \textbf{(D)}\ 3 : 10\qquad \textbf{(E)}\ 1...

Nebraska boy posted the following solution:

he means vertex C

$ b^2 \equal{} s.a \ , \ a^2 \equal{} r.a , therefore r/s \equal{} a^2/b^2 \equal{} 1/9 (B)$

Could someone please explain to me how he came to this?

hello, we have
$ r\equal{}a\cos(\beta)\equal{}\frac{a^2}{c}$
$ s\equal{}b\cos(\alpha)\equal{}\frac{b^2}{c}$
from here we get
$ \frac{r}{s}\equal{}\frac{a^2}{b^2}\equal{}\frac{1}{9}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, we have
$ r \equal{} a\cos(\beta) \equal{} \frac {a^2}{c}$
$ s \equal{} b\cos(\alpha) \equal{} \frac {b^2}{c}$
from here we get
$ \frac {r}{s} \equal{} \frac {a^2}{b^2} \equal{} \frac {1}{9}$.
Sonnhard.[/quote]

thankyou  :lol:


Find the condition of $k,l$ such that for all positive $a,b,c$ we have:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+k\frac{ab+bc+ca}{a^2+b^2+c^2}+3l\frac{ab+bc+ca}{(a+b+c)^2} \ge \frac{3}{2}+k+l$

It's from Vasc idea :

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+k\frac{ab+bc+ca}{a^2+b^2+c^2} \ge \frac{3}{2}+k$

 :)

[quote="AYMANE"]because of the semetry we can suppose that a >= b >= c then 1/(b+c) >= 1/(a+c) >= 1/(a+b);
Tchebytchev : a/(b+c)+b/(a+c)+c/(a+b) >= 1/3.(a+b+c)(1/(b+c)+1/(a+c)+1/(a+b));
Cauchy : ( (b+c)+(a+c)+(a+b)) (1/(b+c)+1/(a+c)+1/(a+b)) >= 9
Then a/(b+c)+b/(a+c)+c/(a+b) >= 3/2.
So if we suppose K and l positive real ( not mentionned ?)
We just need :1) ab +ac + bc >= a 2 +b 2 +c 2
                     2) ab +ac +bc >= 1/3(a 2 +b 2 +c 2 )
which are equivalente to : (a+b+c) 2  >= 3(a 2 +b 2 +c 2 )
orrect me if not[/quote]

1. the question is asking you to find the best constant k and l.
2. Your signs for (1) and (2) are wrong.

I forget writting $k,l$ in the question.

That's why my answer was for an anathor exarcice :D


modular arithmetic
number theory unsolved
number theory

Let $m$ be a positive integer. Prove that there are integers $a, b, k$, such that both $a$ and $b$ are odd, $k\geq0$ and
\[2m=a^{19}+b^{99}+k\cdot2^{1999}\]

This is from China 1999,you can find it in Resources

Yes it's China 1999.
Here I propose the rather more general version : Let be given $m,b\in\mathbb N^*$ where $b$ is odd then there exist $a$ odd and $k\in\mathbb N$ so that $2m=a^{19}+b^{99}+k*2^{1999}$.

[hide="Solution to original"](in collaboration with hyperbolictangent)
It suffices to show that $a^{19}$ and $b^{99}$ range over every odd residue $\bmod{2^{1999}}$ as $a$ and $b$ range over every odd residue $\bmod{2^{1999}}$.

$a^{19}$: Consider $z=(a+t\cdot 2^{1000})^{19}$, where $t\in\mathbb{Z}$. Then $z\equiv a^{19}+19at\cdot 2^{1000}\bmod{2^{1999}}$. Since $\gcd(19a,2^{1999})=1$, we know that as $t$ varies, $19at\cdot 2^{1000}$ takes on every residue $\bmod{2^{1999}}$ that is also a multiple of $2^{1000}$. Hence it suffices to show that $a^{19}$ takes on every odd residue $\bmod{2^{1000}}$. We can continue this argument until we eventually arrive at proving that $a^{19}$ takes on every odd residue $\bmod{4}$, which is clear. This proves that $a^{19}$ takes on every odd residue $\bmod{2^{1999}}$.

$b^{99}$: Consider $w=(b+s\cdot 2^{1000})^{99}$, where $s\in\mathbb{Z}$. Then $w\equiv b^{99}+99bs\cdot 2^{1000}\bmod{2^{1999}}$. Since $\gcd(99b,2^{1999})=1$, we know that as $s$ varies, $99bs\cdot 2^{1000}$ takes on every residue $\bmod{2^{1999}}$ that is also a multiple of $2^{1000}$. Hence it suffices to show that $b^{99}$ takes on every odd residue $\bmod{2^{1000}}$. We can continue this argument until we eventually arrive at proving that $b^{99}$ takes on every odd residue $\bmod{4}$, which is clear. This proves that $b^{99}$ takes on every odd residue $\bmod{2^{1999}}$.

This completes the proof.[/hide]

[hide="Proof of generalization"]This follows quickly; once we have chosen $m$ and $b$, we can choose a sufficiently low (possibly negative) $a$ such that $2m\equiv a^{19}+b^{99}\bmod{2^{1999}}$. Then the choice of a positive $k$ is fixed.[/hide]

Sorry to revive. Take $b=1$. Now we apply Hensel's Lemma on $f(a)=a^{19}+1-2m$ and since $a$ is odd $f(a)\equiv 0\pmod 2$ has a solution and $f'(a)\not\equiv 0\pmod 2$ thus for any $k$ there is $a_k$ with $f(a_k)\equiv 0\pmod{2^{k}}$. In particular we take $k=1999$. And thus $2m=a_{1999}^{19}+1^{99}+k\cdot 2^{1999}$. See we can assume $k\ge 0$ because we can otherwise vary $a_{1999}\mod 2^{1999}$ unless $k\ge 0$.


Because of medical Technology advances the disability rates for people over 65 have been dropping rather dramatically. The function
R(t)= 26.3e^-0.016 (0<=t<=18)
gives the disability rate R(t), in percent, for people over age 65 from 1982 (t=0) through 2000 where t is measured in years.
What was the disability rate in 1986?

I know that your supposed to plug 4 in for t and the problem becomes
26.3e^-.064
and I know e= 2.7182818
but where do I go from here? I don't want the answer I just want to know what I'm doing wrong. Your help would be greatly appreciated.

You are not doing anything wrong!  :)

Just pull out your calculator and evaluate $ 26.3e^{\minus{}0.064}$, and you have your answer.


Find all function $ f: R\rightarrow R$  that satisfy for all $ x,y>0$ : $ f(1\plus{}xf(y))\equal{}yf(x\plus{}y)$ .    (National Contest "Vranceanu Procopiu"-Romania 2004)

Here is a proof if we assume $ f: \mathbb {R}^ \plus{} \rightarrow\mathbb {R}^ \plus{}$.
It can be shortened if we use
"surjective" and "monotonicty" imply continuity.

http://www.mathlinks.ro/viewtopic.php?t=218300


Can I be tied for 7th instead of being in 8th?

lol missing number 1.  I on;y got 2 of the ones I did and I got number 1.

I'm unhappy. I guessed on two of the problems I got wrong, if I 'dlet those be blank I would be several places higher

7 was really easy, but I forgot that 6*1=1*6=6 so my answer was off by 2

[quote="NightFlarer"]Haha Beta believe it or not so did I!
What answer did you pick?
I picked E 19.5[/quote]
17.125

What is an aime-qualifying score for this?

[quote="Rep123max"][b]113[/b]. Genius1601-96[/quote]

When you get a chance, you might want to fix that....should read "13".

[quote="Rep123max"]I think the most missed one was 7, and I don't know if anyone answered #25. Perhaps Teddy did? Did you? I actually think that #25 was easier than some of the others.[/quote]

Yeah, I got #25.  It was moderately easy for #25.

Rep, are you still going to make the graph?

NOOOOOOOO!!! Rosen beat me by 1!!! If I had just gotten 6 right...
I thought 23 was pretty easy though, probably because I didn't really do it, lol.

Rep.  I am 99.9% sure I scored the lowest and it wouldn't bug me a bit.  I am quite proud I scored as high as I did.

[quote="Rep123max"]Is anybody against the graph? Here it is, if you don't want to see it, don't look at it.[/quote]

I am discouraged.  I am suicidal now.
Man, maybe I should work on math stuff again.

You guys are all geniuses.

par excellence!

[quote="nr1337"][quote="h_s_potter2002"]I really don't know what to think about my score.  A lot of people said that Rep's test was more of an AMC 12 level, but if that's true, my score is very very high for me.  But then, I only answered the easy questions.[/quote]

I only answered the easy questions and I still made stupid mistakes!  :D[/quote]

I think it's easier to make stupid mistakes on easy questions since you go through the problem so fast, not really thinking about it, so you're more likely to miss something.

But in hard questions, you look at the question more carefully, so you can understand it better.

Hey guys,

I know i don't post much, but I do follow the board pretty closely. I was planning on taking the mock amc B, but when i woke up yesterday (wed), and tried going to the forum. there was a 400 Result Error. So i figured the AoPS server was having difficulties. 6 o clock came around and I still couldnt get onto the forum, so i figured this was happening to everyone and someone would postpone the mock amc B.

I guess i shouldve read the error because it was a just a stupid cookie on my computer that was preventing me from going into the forum. SO i missed the mock amc B and was bummed out. I took the test this morning and got a 105.5.

dang. tied for 5th.

i know it's unofficial and won't count. but i felt like singing my sorrow.

Rep, for some reason Nosepuppy is in two places on the High Score list...why is that?

[quote="Nosepuppy"]rep- i think there might be a small problem.

i was at a friends house some days ago, and never logged off my account there.  so i think this friend of mine sent in the first set of answers to you.  that's the score of 97.

my real solutions were sent a bit later.

i'm entirely serious.  i know, it's hard to believe, but there are people who know me who are willing to back up my integrity.

sorry for the incovenience =/[/quote]

If you still have the problems can you PM them to me...I accidentally got the calendar mixed up in my head somehow and I didn't answer any of the problems?  If that's so then what is my score? lol, j/p.  Anyways, do you think you could PM me the problems?  Thanks.

Hmm... better be careful with that graph

By manipulating the chart I was able to see everyone's score...

Won't say how, just letting you know though...

[quote="white_horse_king88"]Wait - do you mean you saw everyone's names beside their scores? That's amazing - I have no idea how you did that.[/quote]

I do  :maybe:

Rep, you should delete all the names next to the numbers, the graph will still look the same

[quote="Philip_Leszczynski"]Ahhh! I would have been tied at 131 with Teddy if only I partook![/quote]

 :| It's a mock contest...I hardly think it's worth reviving an old topic like this just to say that.


A party has $ 6$ attendees. Attendees shake hands with other attendees, some pairs may shake hands more than once or not at all. Each list below records the number of hands each attendee shook at the party. Which one of the lists must be incorrect?

$ \mathrm{(A)}\ (1,\,2,\,2,\,2,\,2,\,3)\qquad\mathrm{(B)}\ (1,\,3,\,3,\,3,\,4,\,4)\qquad\mathrm{(C)}\ (3,\,3,\,3,\,3,\,3,\,3)\qquad\mathrm{(D)}\ (2,\,2,\,3,\,3,\,3,\,4)\qquad\mathrm{(E)}\ \text{None of these}$

[hide]When people shake hands, there must be an even number of total shakes from each side, since $ 2$ people are involved in $ 1$ handshake. Only choice $ \fbox{D}$ has an odd number of shakes: $ 2\plus{}2\plus{}3\plus{}3\plus{}3\plus{}4\equal{}17$ handshakes. [/hide]


conics
parabola
probability
trigonometry
quadratics
algebra
binomial theorem

11.) The parabola $y = x^2$ is reflected in the line y = 3, producing a new parabola. This new parabola is reflected in the line x = 2, producing the parabola $y = -x^2 + bx + c$. Compute the ordered pair (b,c).

12.) Find the largest value of x for which $x^2 + y^2 = x + y$ has a solution, if x and y are real.

13.) Let z be such that $z^7 = 1$ and $z \neq 1$. Compute the numerical value of $z^{10} + \frac{1}{z^{10}} + z^{30} + \frac{1}{z^{30}} + z^{50} + \frac{1}{z^{50}}$

14.) Find, as a proper base ten fraction, the sum of the following infinite series of repeating decimals, all written in base eight:
S = .07* + .007* + .0007* + ... (where * indicates a repeating digit)

15.) Find all positive integer values of n, 1 < n < 100, such that $\binom{n}{2}$ is a perfect square.

16.) An “unfair” coin has a 2/3 probability of turning up heads. If this coin is tossed 50 times, what is the probability that the total number of heads is even?

17.) Find the smallest positive integer such that when its leading digit is removed and placed at the end of the number, the resulting number is twice as large as the original.

18.) Find the value of $\binom{50}{50} - \binom{50}{48} + \binom{50}{46} - \ldots - \binom{50}{4} + \binom{50}{2} - \binom{50}{0}$

19.) Given that $1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots = \frac{\pi^2}{6}$, find the value of $1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots$

[hide="Number 18"]
Consider $(1+i)^{50}$.  Using the binomial theorem, we can expand this as

\begin{eqnarray*} (1+i)^{50} &=& \binom{50}{50}+\binom{50}{49}i+\binom{50}{48}i^2+\binom{50}{47}i^3+\cdots+\binom{50}{1}i^{49}+\binom{50}{0}i^{50}\\ &=& \binom{50}{50}+\binom{50}{49}i-\binom{50}{48}-\binom{50}{47}i+\cdots+\binom{50}{1}i-\binom{50}{0} \end{eqnarray*}

where every other term has an $i$ and the sign of the terms switch every two terms.  So then we can write this as \[ \left[\binom{50}{50}-\binom{50}{48}+\cdots+\binom{50}{2}-\binom{50}{0}\right]+\left[\binom{50}{49}-\binom{50}{47}+\cdots-\binom{50}{3}+\binom{50}{1}\right]i. \]

Basically we just showed that what we are trying to find is the real part of $(1+i)^{50}.$

We can convert $(1+i)^{50}$ to polar form using $1+i=\sqrt{2}\text{cis } 45^\circ.$  Thus

\begin{eqnarray*} (1+i)^{50} &=& (\sqrt{2}\text{cis }45^\circ)^{50}\\ &=& (\sqrt{2})^{45}\text{cis }(45\cdot 50)^\circ\\ &=& 2^{45/2}\text{cis } 2250^\circ\\ &=& 2^{45/2}\text{cis } 90^\circ\\ &=& 2^{45/2}\cos 90^\circ + 2^{45/2}i\sin 90^\circ\\ &=& 0 + 2^{45/2}i\end{eqnarray*}

Thus we can match up real and imaginary components:

\begin{eqnarray*} 0+2^{45/2}i &=& \left[\binom{50}{50}-\binom{50}{48}+\cdots+\binom{50}{2}-\binom{50}{0}\right]+\left[\binom{50}{49}-\binom{50}{47}+\cdots-\binom{50}{3}+\binom{50}{1}\right]i\\ &\Rightarrow & \binom{50}{50}-\binom{50}{48}+\cdots+\binom{50}{2}-\binom{50}{0} = 0\\ &\text{and}& \binom{50}{49}-\binom{50}{47}+\cdots -\binom{50}{3}+\binom{50}{1} = 2^{45/2} \end{eqnarray*}

So our answer is $\boxed{0.}$[/hide]

Hints to the ones i did:

[hide]12. Kinda reason through, manipulating takes more effort and time

13. Look at $\mod{7}$

14. I think it's 1/7...not sure =(

15. $\frac{n(n-1)}{2}=k^2$, now look at divisibility by $2$ and trial and error because we need the other factor and the quotient to be a square.

17. $199a=80b+8c$ where the original number is $abc$
[/hide]

[quote="gauss202"]I would prefer you put your answers. That way I can tell you if you're right or not. Your hints are kind of vague.[/quote]

OK, i'll post my solutions tonight, I'm a bit busy right now. So far numbers 11,19,16, and 8 have been bugging me :(

[hide="Number 11"]
Basically, reflecting $y=x^2$ over $y=3$ does two things.  First, it flips $y=x^2$ over and, second, it translates it upwards 6 units.  So now the vertex of the parabola is at $(0,6)$.  It follows that the new equation is $y=-x^2+6.$

Reflecting this new parabola over $x=2$ does not change which way the parabola is facing.  It just translates the parabola over 4 units, essentially.  So the vertex is now at $(4,6)$. Thus we have $y=-(x-4)^2+6=-x^2+8x-10\Rightarrow (b,c)=\boxed{(8,-10).}$[/hide]

[hide="Number 16"]
For the number of heads to be even, we can have 0 heads, 2 heads, 4 heads, and so forth, up to 50 heads.  Now we want to add up the probability of each one of these cases.  First, we need to find the probability of each one.  If we have $k$ heads, then we must have $50-k$ tails.  There are $\binom{50}{k}$ ways to get $k$ heads when tossing the coins.  Thus our probability of getting $k$ heads is $\binom{50}{k}\left(\frac 23\right)^k\left(\frac 13\right)^{50-k}.$  We want to sum this for $k=0,2,4,\cdots,50.$  Thus we want

\[ \binom{50}{0}\left(\frac 23\right)^0\left(\frac 13\right)^{50}+\binom{50}{2}\left(\frac 23\right)^2\left(\frac 13\right)^{48}+\cdots+\binom{50}{50}\left(\frac 23\right)^{50}\left(\frac 13\right)^{0}=\frac 1{3^{50}} \left[\binom{50}{0}+2^2\binom{50}{2}+\cdots+2^{50}\binom{50}{50}\right]. \]

That sum looks kinda hard to evaluate.  But it does remind of us some things which we can sum easily.  From the binomial theorem with $(x+y)^{50}$, $(x-y)^{50}$, $x=2$ and $y=1$, we get:

\begin{eqnarray*} \binom{50}{0}+2\binom{50}{1}+2^2\binom{50}{2}+\cdots+2^{50}\binom{50}{50} &=& 3^{50}\\ \text{and} \binom{50}{0}-2\binom{50}{1}+2^2\binom{50}{2}-\cdots+2^{50}\binom{50}{50} &=& 1 \end{eqnarray*}

Adding these two yields $2\left[\binom{50}{0}+2^2\binom{50}{2}+\cdots+2^{50}\binom{50}{50}\right] = 3^{50}+1.$

It follows that our answer is thus $\boxed{\frac{3^{50}+1}{2\cdot3^{50}}}.$

You can generalize this to $\frac{3^n+1}{2\cdot3^n}$ where $n$ is the number of coins.[/hide]

[hide="Number 19"]
Let $S=1-\frac 1{2^2}+\frac 1{3^2}-\cdots$ and $T=1+\frac 1{2^2}+\frac 1{3^2}+\cdots$.  Then \[ S+T = 2\left(1+\frac 1{3^2}+\frac 1{5^2}+\cdots\right). \]

We can evaluate the RHS of the above equation(leave off the 2 out in front of the parentheses for now):

\begin{eqnarray*} 1+\frac 1{3^2}+\frac 1{5^2}+\cdots &=& \left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right) - \left(\frac 1{2^2}+\frac 1{4^2}+\frac 1{6^2}+\cdots\right)\\ &=& \left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)-\frac 1{2^2}\left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)\\ &=& \left(1-\frac 1{2^2}\right)\left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)\\ &=& \frac 34\cdot \frac{\pi^2}6\\ &=& \frac{\pi^2}8 \end{eqnarray*}

So

\begin{eqnarray*} S+T &=& 2\left(1+\frac 1{3^2}+\frac 1{5^2}+\cdots\right)\\ S+\frac{\pi^2}6 &=& 2\left(\frac{\pi^2}8\right)\\ \Rightarrow S &=& \boxed{\frac{\pi^2}{12}} \end{eqnarray*}[/hide]

heh, looks like you've already started. I guess I'll just post solutions one by one then.
[hide="13"]
After rationalizing the denominator, bringing it to one single fraction, and cancelling out $z^7$ (since it equals $1$), we have that the sum we are looking for is $z^6+z^5+z^4+z^3+z^2+z$, let us call this $S$. Now notice that $z^7-1=(z-1)(z^6+z^5+z^4+z^3+z^2+z+1)=(z-1)(S+1)=0$. Now, since $z \not=1$, then $S+1$ must $=0$, therefore $S=\boxed{-1}$
[/hide]

[quote="Iversonfan2005"]heh, looks like you've already started. I guess I'll just post solutions one by one then.[/quote]

Posting 3 at a time is far better.  Look at joml88's post, for example.

[quote="3"]Posting 3 at a time is far better.[/quote]

Ha, ha, ha...  :rotfl:

3...

[quote="joml88"]

[hide="Number 19"]
Let $S=1-\frac 1{2^2}+\frac 1{3^2}-\cdots$ and $T=1+\frac 1{2^2}+\frac 1{3^2}+\cdots$.  Then \[ S+T = 2\left(1+\frac 1{3^2}+\frac 1{5^2}+\cdots\right).  \]

We can evaluate the RHS of the above equation(leave off the 2 out in front of the parentheses for now):

\begin{eqnarray*} 1+\frac 1{3^2}+\frac 1{5^2}+\cdots &=& \left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right) - \left(\frac 1{2^2}+\frac 1{4^2}+\frac 1{6^2}+\cdots\right)\\ &=& \left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)-\frac 1{2^2}\left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)\\ &=& \left(1-\frac 1{2^2}\right)\left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)\\ &=& \frac 34\cdot \frac{\pi^2}6\\ &=& \frac{\pi^2}8 \end{eqnarray*}

So

\begin{eqnarray*} S+T &=& 2\left(1+\frac 1{3^2}+\frac 1{5^2}+\cdots\right)\\ S+\frac{\pi^2}6 &=& 2\left(\frac{\pi^2}8\right)\\ \Rightarrow S &=& \boxed{\frac{\pi^2}{12}} \end{eqnarray*}[/hide][/quote]
Hmm... easier way?

\begin{eqnarray*} \frac{\pi^2}6 &=& 1+\frac 1{2^2}+\frac 1{3^2}+\frac1{4^2}+\cdots \\ S &=& 1-\frac 1{2^2}+\frac 1{3^2}-\frac1{4^2}+\cdots \\ \text{Subtract: } \frac{\pi^2}6-S &=& \frac 2{2^2}+\frac 2{4^2}+\frac2{6^2}+\cdots=\frac12\left(1+\frac 1{2^2}+\frac 1{3^2}+\cdots\right)=\frac{\pi^2}{12} \\ S &=& \frac{\pi^2}6-\frac{\pi^2}{12}=\boxed{\frac{\pi^2}{12}} \end{eqnarray*}

[quote="gauss202"]12.) Find the largest value of x for which $x^2 + y^2 = x + y$ has a solution, if x and y are real.[/quote]

[hide="12"]
$x^2-x+(y^2-y)=0$

$x=\frac{1\pm\sqrt{1-4(y^2-y)}}{2}$

$x=\frac{1\pm\sqrt{-4y^2+4y+1}}{2}$

$\therefore x\in\mathbb{R}\iff -4y^2+4y+1\ge 0$

$\iff \frac{1}{2}-\frac{\sqrt{2}}{2}\le y\le\frac{1}{2}+\frac{\sqrt{2}}{2}$

$\Rightarrow \boxed {\frac{1}{2}+\frac{\sqrt{2}}{2}}$
[/hide]

[quote="gauss202"]15.) Find all positive integer values of n, 1 < n < 100, such that $\binom{n}{2}$ is a perfect square.[/quote]

[hide]$n^2-n=2k^2$

The discrimant of this quadratic equation is $8k^2+1$. If we want $n\in\mathbb{Z}$, then $8k^2+1=z^2$ for $z\in\mathbb{Z}$.

$\iff z^2-8k^2=1$

PELL!

$\sqrt{8}=2+\frac{1}{1+\frac{1}{4+\frac{1}{1+\frac{1}{4+\cdots}}}}$

convergents: $\frac{2}{1}$, $\frac{3}{1}$, $\frac{14}{5}$, $\frac{17}{6}$, $\frac{82}{29}$, $\frac{99}{35}$, ... (the rest are too big)

period is 2

$(z,k)=(3,1),(17,6),(99,35),\ldots$

$n=\frac{1\pm z}{2}$

$=\boxed {2,9,50}$[/hide]


[quote="jin thynj"]How many 4-tuples (a, b, c, d) satisfy 0 < a < b < c < d < 500, a + d = b + c, and bc - ad = 93?[/quote]

Let $ a\equal{}b\minus{}k$ , then $ d\equal{}c\plus{}k$, where $ k$ is positive integer.

So $ bc\minus{}ad\equal{}93 \Leftrightarrow bc\minus{}(b\minus{}k)(c\plus{}k)\equal{}93$
$ \Leftrightarrow k(c\minus{}b\plus{}k)\equal{}93\equal{}3 \times 31$

$ k|93$ and $ k^2<k(c\minus{}b\plus{}k)\equal{}93$
$ \Rightarrow$ Possible values of $ k$ are $ 1,3$

$ k\equal{}1 \Rightarrow c\minus{}b\equal{}92 \Rightarrow (a,b,c,d)\equal{}(k,k\plus{}1,k\plus{}93,k\plus{}94)$, where $ 1 \le k \le 405$ as solutions.
$ k\equal{}3 \Rightarrow c\minus{}b\equal{}28 \Rightarrow (a,b,c,d)\equal{}(m,m\plus{}3,m\plus{}31,m\plus{}34)$, where $ 1 \le m \le 465$ as solutions.

So the answer $ \equal{}405\plus{}465\equal{}870$

[quote="jin thynj"]Perfect. As xpmath said you can look also [url=http://www.mathlinks.ro/viewtopic.php?p=461985#461985]here[/url]  :wink:[/quote]

If you know that there is another thread dedicated to each problem, why make another thread? Maybe you realized it afterwards...

Regardless, please use Contests section to see the previous years' AIME problems.

Locked.


quadratics
modular arithmetic
algebra
polynomial
calculus
derivative

2 questions:

How can you tell if a residue class is a quadratic residue mod $ p^n$ for some prime $ p$ and natural number $ n$?  Please prove.
(The article I am reading ("Quadratic Congruences" by Dusan Djukic, available at [url]http://www.imomath.com[/url]) says for $ a$ to be a quadratic residue mod $ p^n$, either it is a quadratic residue mod $ p$ or $ p^2|a$ and $ a/p^2$ is a quadratic residue mod $ p^{n\minus{}2}$.  I don't understand the first part: why $ a$ needs be a quadratic residue mod $ p$ for it to be a quadratic residue mod $ p^n$ when $ p^2$ does not divide $ a$)

And somewhat related: why are the numbers $ a \equiv 1 \pmod{8}$ quadratic residues mod $ 2^n$, where $ n \ge 3$?  And why are these the only odd quadratic residues in mod $ 2^n$?

1)  $ a$ is a quadratic residue $ \bmod p^n$ if and only if the polynomial $ x^2 \minus{} a \equiv 0 \bmod p^n$ has a solution.  Clearly $ x^2 \minus{} a \equiv 0 \bmod p$ is a necessary condition.  As for sufficient conditions, apply [url=http://en.wikipedia.org/wiki/Hensel%27s_lemma]Hensel's lemma[/url].

2)  The second part follows from the first part.  The first part is a consequence of the fact that $ \mathbb{Z}/2^n \mathbb{Z}$ doesn't have a primitive root for $ n \ge 3$; the element of greatest order has order $ 2^{n\minus{}2}$.

1) Hensel's lemma definitely solves the problem, but I'd like to see a proof (at least for the case where $ f(x)\equal{}x^2$)

2) lost... I realize it is necessary that $ a \equiv 1 \pmod{8}$ for $ a$ to be an odd quadratic residue mod $ 2^n$ but haven't figured out how to prove it is sufficient.

The basic idea of Hensel's lemma is as follows:  if $ f(a) \equiv 0 \bmod p$ then a solution $ \bmod p^2$ must reduce to a solution $ \bmod p$, so any such solution must be of the form $ f(a \plus{} tp) \equiv 0 \bmod p^2$ for some $ t$.  If $ f'(a) \neq 0$ then

$ f(a \plus{} tp) \equiv f(a) \plus{} pt f'(a) \bmod p^2 \Leftrightarrow$
$ \frac{f(a)}{p} \plus{} t f'(a) \equiv 0 \bmod p$

and it is possible to find a unique such $ t$; otherwise, all such $ t$ work.  (The second case gives you the answer to your second question.)

Let $ f(x) \equal{} x^2 \minus{} c$.  Then $ f'(x) \equiv 2x \pmod{p}$ so the derivative is only equivalent to $ 0 \pmod{p}$ when $ p|x$ OR when $ p \equal{} 2$.  So the powers of 2 are problematic.

Hensel's Lemma then claims that if $ f(r) \equiv 0 \pmod{p^{k}}$, then $ f(x) \equiv 0 \pmod{p^{k}}$ for any $ x \equiv r \pmod{p^{k \minus{} 1}}$.  That is, if we can find a solution $ r$ to $ x^2 \minus{} c \equiv 0 \pmod{2^n}$ then $ r \plus{} 2^{n \minus{} 1}$ is also a solution.  So for each odd $ c$, if we have a solution we will have exactly $ 4$ solutions $ \pm r$ and $ \pm(r \plus{} 2^{n \minus{} 1})$ - as long as $ n \ge 3$.  Of course, all these solutions must be odd.

These $ 4$ solutions for each odd quadratic residue implies that 1/4 of all odd residue classes are quadratic residues.  $ x^2 \minus{} c \equiv 0 \pmod{2^n}$ implies that $ x^2 \minus{} c \equiv 0 \pmod{8}$, so $ c$ can only be a quadratic residue mod $ 2^n$ if it is a quadratic residue mod $ 8$.  Therefore, it is necessary that $ c \equiv 1 \pmod{8}$.  Now the claim that exactly 1/4 of all odd residue classes are quadratic implies that every $ c\equiv 1 \pmod{8}$ must be a quadratic residue.

I suppose this proves the claim.  I thought I would be asking for you to fill in some parts of this proof.

Anyway, I still have 2 questions:
1)where does the congruence $ f(a \plus{} tp^{k \minus{} 1}) \equiv f(a) \plus{} tp^{k \minus{} 1}f'(a) \pmod{p^k}$ come from? (Note $ f(a) \equiv 0 \pmod{p^{k \minus{} 1}}$ is given).  The whole theorem seems to be built around this congruence.

2) what does the fact that $ 2^n$ has no primitive roots have to do with it?  Also, how do you prove that the element with greatest order has order $ 2^{n \minus{} 2}$?  I've always been a bit curious about this since I learned about Carmichael's theorem.

EDIT: actually, I suppose my proof is incomplete.  I showed that there are at least 4 roots; I did not rule out the possibility of there being more, and then some $ c$ not bein quadratic residues as a result.  In mod $ 2^{n\plus{}1}$, the residue class $ c \pmod{2^n}$ is split between $ c \pmod{2^{n\plus{}1}}$ and $ c\plus{}2^n \pmod{2^{n\plus{}1}}$.  It is enough to observe that $ r^2 \neq (r\plus{}2^{n\minus{}1})^2 \equiv r^2 \plus{}2^n r \pmod{2^{n\plus{}1}}$ so $ \pm r$ and $ \pm r\plus{}2^{n\minus{}1}$ are solutions to $ x^2\minus{}c \equiv 0 \pmod{2^{n\plus{}1}}$ and $ x^2\minus{}(c\plus{}2^n) \equiv 0 \pmod{2^{n\plus{}1}}$ in some order.

Specifically, when the $ k$ in $ 2^{n}k \equal{} r^2\minus{}c$ is even, $ \pm r$ are the solutions to $ x^2\minus{}c \equiv 0 \pmod{2^{n\plus{}1}}$ and $ r^2\plus{}2^n(r\minus{}1)\minus{}c \equiv 0 \pmod{2^{n\plus{}1}}$ must hold since $ r\minus{}1$ is even, implying $ x\equal{}\pm(r\plus{}2^{n\minus{}1})$ is a solution to $ x^2\minus{}c\minus{}2^{n} \equiv 0 \pmod{2^{n\plus{}1}}$; when $ k\equal{}2k'\minus{}1$ is odd, $ 2^{n}(2k'\minus{}1) \equal{} 2^{n\plus{}1}k' \minus{} 2^{n} \equal{} r^2\minus{}c\minus{}2^n$ so $ \pm r$ would clearly be a solution to $ x^2\minus{}c \minus{}2^n\equiv 0 \pmod{2^{n\plus{}1}}$, and $ x\equal{}\pm (r\plus{}2^{n\minus{}1})$ is clearly a solution to $ x^2\minus{}c \equiv 0 \pmod{2^{n\plus{}1}}$.  We again reach $ 4$ solutions for each congruence by adding $ 2^n$ to each of our solutions.  And we can finish by noting that there are only $ 4$ solutions to $ x^2 \equiv 1 \pmod{8}$ so by induction, each time we go to the next power of 2 ($ 2^{n\plus{}1}$), we cut our number of solutions in half but then double our number of solutions by finding new ones by adding $ 2^n$ to each existing solutions, so each time we always have found $ 4$ solutions for each $ c\equiv 1 \pmod{8}$, which means we have counted exactly as many solutions to $ x^2 \minus{} c \equiv 0$ as their are odds available for $ x$.

[quote="facis"]1)where does the congruence $ f(a \plus{} tp^{k \minus{} 1}) \equiv f(a) \plus{} tp^{k \minus{} 1}f'(a) \pmod{p^k}$ come from? (Note $ f(a) \equiv 0 \pmod{p^{k \minus{} 1}}$ is given).  The whole theorem seems to be built around this congruence.[/quote]
$ f$ is a polynomial.  Expand the terms and remember that any time $ p^{k\minus{}1}$ is taken to higher than a first power it vanishes $ \bmod p^k$.  See also the comments [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=225324]here[/url].

[quote="facis"]2) what does the fact that $ 2^n$ has no primitive roots have to do with it?  Also, how do you prove that the element with greatest order has order $ 2^{n \minus{} 2}$?  I've always been a bit curious about this since I learned about Carmichael's theorem.[/quote]
The standard proof I know exhibits it:  I believe $ \minus{}5$ is such an element for every $ n \ge 3$ (that might actually be $ 5$).  The proof then proceeds by the binomial theorem.  There is probably a nicer way to see this, but the point is that you then know that $ \mathbb{Z}/(2^n) \simeq C_2 \times C_{2^{n\minus{}2}}$, and we know how to count squares in an abelian group.

[quote="t0rajir0u"]The standard proof I know exhibits it:  I believe $ \minus{} 5$ is such an element for every $ n \ge 3$ (that might actually be $ 5$).  The proof then proceeds by the binomial theorem.  There is probably a nicer way to see this, but the point is that you then know that $ \mathbb{Z}/(2^n) \simeq C_2 \times C_{2^{n \minus{} 2}}$, and we know how to count squares in an abelian group.[/quote]

I actually know very little about groups, and I have no clue what you mean by $ \mathbb{Z}/(2^n) \simeq C_2 \times C_{2^{n \minus{} 2}}$.  Can you please explain?

If $ q$ has a primitive root, then $ (\mathbb{Z}/(q))^{*}$ is, multiplicatively, a cyclic group - that is, it looks like $ (\mathbb{Z}/(\varphi(q))^{\plus{}}$ with the homomorphism given by $ n \mapsto g^n$ where $ g$ is a primitive root.

On the other hand, $ (\mathbb{Z}/(2^n))^{*}$ doesn't have this property; it has $ 2^{n\minus{}1}$ elements but the maximal order of an element is $ 2^{n\minus{}2}$.  So instead of being a cyclic group it is a [url=http://en.wikipedia.org/wiki/Direct_product]direct product[/url] of two cyclic groups:  it behaves like the set of tuples $ (a, b), a \in (\mathbb{Z}/(2))^{\plus{}}, b \in (\mathbb{Z}/(2^{n\minus{}2}))^{\plus{}}$ under componentwise addition.

Can I clarify, is $ \mathbb{Z}/(2^n)$ the same as $ \mathbb{Z}/2^n\mathbb{Z}$? (I haven't seen that notation before)

Yes; the notation comes from ring theory.  $ (p)$ means "the [url=http://en.wikipedia.org/wiki/Principal_ideal]principal ideal[/url] generated by $ p$."  It is usually not used in connection to modular arithmetic, but $ \mathbb{Z}/p\mathbb{Z}$ is more lattice-theoretic than ring-theoretic and $ \mathbb{Z}_p$ also refers to the $ p$-adic integers.

That's odd, my textbook uses triangular type brackets (eg $ \langle a \rangle$) for ideals. Is this non-standard?

Artin and Wikipedia both use parentheses, but I can see why you wouldn't want to; the notation is somewhat ambiguous, although it is a useful ambiguity when the notation $ (a, b)$ can be interpreted both as the ideal generated by $ a, b$ and the greatest common divisor of $ a, b$; over a principal ideal domain, the two notions coincide.

Could you please explain what all this notation means?  I think I get some of what you are trying to say, but not all.

[quote="t0rajir0u"]If $ q$ has a primitive root, then $ (\mathbb{Z}/(q))^{*}$ is, multiplicatively, a cyclic group - that is, it looks like $ (\mathbb{Z}/(\varphi(q))^{ \plus{} }$ with the homomorphism given by $ n \mapsto g^n$ where $ g$ is a primitive root.
[/quote]

I think I get this.  You are saying that the elements in mod $ q$ relatively prime to $ q$ can all be uniquely generated by doing $ g^n \pmod q$, where $ g$ is a primitive root and $ 0 \le n < \phi(q)$.  What I don't get is what your $ *$ and $ \plus{}$ mean, and if there is anything special meant by using $ \varphi(q)$ instead of $ \phi(q)$.

[quote="t0rajir0u"]
On the other hand, $ (\mathbb{Z}/(2^n))^{*}$ doesn't have this property; it has $ 2^{n \minus{} 1}$ elements but the maximal order of an element is $ 2^{n \minus{} 2}$.  So instead of being a cyclic group it is a [url=http://en.wikipedia.org/wiki/Direct_product]direct product[/url] of two cyclic groups:  it behaves like the set of tuples $ (a, b), a \in (\mathbb{Z}/(2))^{ \plus{} }, b \in (\mathbb{Z}/(2^{n \minus{} 2}))^{ \plus{} }$ under componentwise addition.[/quote]

More lost here... with the direct product, are you trying to say all elements relatively prime to $ 2^n$ can be represented as $ 4b\plus{}2a\plus{}1$?  I understand when you say it isn't a cyclic group, but what exactly does the direct product mean?

1.  $ (\mathbb{Z}/(q))^{*}$ means "the group of units in $ \mathbb{Z}/(q)$," i.e. the group of invertible elements under multiplication.  $ \varphi(q)$ is the same as $ \phi(q)$.  $ (\mathbb{Z}/(\varphi(q))^{ \plus{} }$ means "the group $ \mathbb{Z}/(\varphi(q))$ under addition."  The reason this is necessary is that $ \mathbb{Z}/(p)$ is a ring, so it has compatible addition and multiplication and when talking about groups it is necessary to specify which one you're talking about.

There is an abstract structure here, the [i]cyclic group[/i], which multiplicatively looks like $ \{ 1, g, g^2, ... g^{n \minus{} 1} \}$ where $ g$ is a generator and $ n$ is the order of the group, and additively looks like $ \{ 0, 1, 2, ... n \minus{} 1 \}$ where $ 1$ is the generator.

2.  Perhaps this would be clearer with an example.  Take $ (\mathbb{Z}/(8))^{*}$:  its elements are $ \{ 1, 3, 5, 7 \}$ and each element has order $ 2$, so it is not a group of the form $ \{ 1, g, g^2, g^3 \}$.  Instead, abstractly it is of the form $ \{ 1, a, b, ab \}$ (where you can take $ a, b$ to be any pair of the three non-identity elements):  this is a group known as the [url=http://en.wikipedia.org/wiki/Klein_four-group]Klein four-group[/url].

It can be represented by a direct product as follows:  I claim that this group is isomorphic to  $ C_2 \times C_2$, the direct product of two cyclic groups of order $ 2$, which is the set $ \{ (0, 0), (1, 0), (0, 1), (1, 1) \}$ with componentwise addition $ \bmod 2$.  Thus if we take

$ (1, 0) \mapsto 3$
$ (0, 1) \mapsto 5$

then

$ (1, 1) \equal{} (1, 0) \plus{} (0, 1) \mapsto 3 \cdot 5 \equal{} 7$

and every element has order $ 2$.


hey evry1
i cant decide which category this problem should  go to (it's left to valentin).here's the problem
1)A rational point in the plane is one with rational coordinates. Divide the rational points into three disjoint sets so that every line contains points from at most two sets, but every circle with rational center contains points from all three sets.
i am not even able to begin on this one
so sum1 help.
regards

well the moving was easy ... let's try to solve it ... this should prove more challenging :)

i suppose the by default the problem is indeed correct, and we CAN divide the rational points into 3 subsets with the required property?

John Sholes doesn't know how to solve it either:

[url]http://www.kalva.demon.co.uk/china/chi02.html[/url]

but the problem statement seems at least to be correct :)

Today, an attempt appeared on John Sholes' site:

[i]We may write a rational point as (a/b, c/d) with a, b, c, d integers, where a and b are coprime and c and d are coprime. Let A be the set of points (a/b, c/d) with b and d odd, B be the set of points (a/b, c/d) with a and d odd and b and d even, or a and d even and b and d odd, let C be the remaining points. We show first that no line contains a point from A and a point from B and a point from C. [/i]

That's all that he put online for the moment.

[quote=wpolly]For any rational line (line with at least two rational points on it) we can write its equation as Ax+By+C=0 with A, B, C are integers and are relatively prime.
Divide the lines into three categories:
1) A,B even, C odd;
2) One of A and B is even, the other is odd;
3) A and B are both odd.
We first show that there is no point which is contained by all 3 kinds of lines.
Suppose there is such a point, let the equations of them are A1x+B1y+c1=0 and so on. Then we have A1, B1, B2(or A2) are even, C1, A2(or B2), A3, B3 are odd.
Thus [code]
  |A1 B1 C1|         |0 0 1 |
0=|A2 B2 C2|    =    |0 1 C2|  =1  (mod 2)
  |A3 B3 C3|         |1 1 C3|
[/code]
Contradiction.
Now we define set A1 is the points which lies on no first-kind lines, and A2, A3 similarly.
Clearly they satisfy the condition (1).
To show that they satisfy condition (2) is tricky, but I believe somebody here could work it out.[/quote]

wow code tags worked really different back then


geometry
calculus
integration
real analysis
real analysis unsolved

what is the antiderivative for y=x^x,and find its area from 1 to 60
find the area of 1/x from 5 to infinite, there should be a value because 1/0=infinite and 1/infinte=0

(a) Not an elementary function.

(b) The improper integral diverges.


quadratics
number theory unsolved
number theory

Find the maximum value of positive integer $n$ so that the system has an integer solution:
$(x + 1)^2 + (y_1)^2 = (x + 2)^2 + (y_2)^2 = ... = (x + n)^2 + (y_n)^2$.

I remember [b]heartwork[/b] sent the general problem some time ago: he asked to find maximal $n$ s.t. there exists quadratic trinomial $P(x)\in\mathbb{Z}[x]$ s.t. $P(1)$, ..., $P(n)$ are perfect squares.

[quote="RaduSorici"]Have a look at this!
http://www.mathlinks.ro/Forum/viewtopic.php?t=29433[/quote]
Thanks a lot!


induction
inequalities
Gauss
pigeonhole principle

Please note that this thread is only for proofs, not for thread where you can post random problems.

I'm making this thread to show beginners about what the "proof" is.

Prove that $1+2+3+.....n = \frac {n(n+1)}{2}$

[hide="Proof"]

This proof can be done by MMI or Method of Mathematical Induction. In induction, we start off by proving when n=1, then assume that n satisifies the problem and prove on case of n+1.

Always, picture is better than words.

$1+2+3+....n = \frac {n(n+1}{2}\\
\text {Test on when n=1}\\
1 = \frac {1(2)}{2} = 1\\
\text {So, proved on 1!}\\
(1+2+3....n)+(n+1) = \frac {(n+1)(n+2)}{2}\\
\frac {n(n+1)}{2} + \frac {2n+2}{2} = \frac {n^2+3n+2}{2} \text {This is from Left Side}\\
\text {Right Side}\\
\frac  {n^2+3n+2}{2}\\
\text {So, it's same on both sides. Thus, this is proven.}$[/hide]

Interesting, isn't it? :)

Here is another famous proof on AM-GM inequality.

AM = Arithmetic Mean (average)
GM = Geometric Mean

AM-GM inequality states this:

$\frac {a+b}{2} \geq \sqrt {ab}$

Now, how do we prove this?

I haven't done this a lot times so there might be other ways but here is what I would do:

[hide="Inequality1"]

First, let's assume $a=b$ which is possible.

Then:

$\frac {2b}{2} = b\\
\text {For the right side}\\
\sqrt {b^2} = b$

Thus, the equal sign is possible. Now, we have to prove in which $a \neq b$ and AM > GM. Here, we'll use something called contradiction which disapproves the first assumption.

Let's assume $\frac {a+b}{2} < \sqrt {ab}$.

Then we can do:

$\frac {a+b}{2} < \sqrt {ab}\\
a+b < 2 \sqrt {ab}\\
a^2+2ab+b^2 < 4ab\\
a^2+b^2 < 2ab$

Let's use our simplest case, 1 and 2. When we plug a=1 and b=2 (doesn't matter which order you use), we get:

$1^2+2^2 = 5\\
2 \cdot 2 \cdot 1 = 4\\
\text {But}\\
5 < 4$

So, our assumption is wrong and that $\frac {a+b}{2} > \sqrt {ab}$. And by the first proof we did, we can now summarize our AM-GM proof as:

$\frac {a+b}{2} \geq \sqrt {ab}$.[/hide]

I'm making this post because I want people to see examples on how proofs should look like:

[hide=""Proof by induction""]Statement: $1+2+\cdots+n=\frac{n(n+1)}{2}$

Assume that $n=1$. Obviously $1=\frac{(1)\cdot(1+1)}{2}=1$.

Suppose the statement is true for $n$ and we can safely assume that $1+2+\cdots+n=\frac{n(n+1)}{2}$ holds.

Then, $1+2+\cdots+n+(n+1)=\frac{n(n+1)}{2}+(n+1)=(n+1)\left(\frac{n}{2}+1\right)=\frac{(n+1)(n+2)}{2}$

Therefore, the statement also holds for $n+1$

By induction hypothesis, the statement is true for all $n\geq 1$.[/hide]

[hide=""Proof of basic 2 variable AM-GM""]Proving inequalities are the best proofs you can do. It takes cleverness on how you manipulate the variables and how you 'see through' a sense of measure in an inequality. This may sound deep and confusing, but my opinion will make some sense once you have proven many inequalities. Here we got a classical inequality that by working forward and simplifying until you see something obvious will be very rewarding.

So how can you prove $\frac{a+b}{2}\geq \sqrt{ab}$ when both $a,b$ are positive numbers? Note that the condition that $a,b$ are positive is [u]necessary[/u].

Let's try to work on something familiar: so the above inequality is equivalent to $a+b\geq 2\sqrt{ab}$.

Let's see what happens when we square both sides; this is a valid operation since we are working with positive numbers: $(a+b)^2\geq 4ab$

But, $(a+b)^2=a^2+2ab+b^2\geq 4ab$ and if we try to simplify we get $a^2-2ab+b^2\geq 0$.

Recall that $a^2-2ab+b^2=(a-b)^2$ and so we have $(a-b)^2\geq 0$. The last part is true since the square of any real number is always greater than or equal to 0.

Now, if you work backwards using the fact that the square of any real number is always greater than or equal to 0, we can form the following steps:

$(a-b)^2\geq 0$ (Initial statement)

$a^2-2ab+b^2\geq 0$ (Expand square)

$a^2+2ab+b^2\geq 4ab$ (Add 4ab to both sides, a positive quantity, preserves inequality)

$(a+b)^2\geq 4ab$ (Reduce to square form)

$a+b\geq 2\sqrt{ab}$ (Taking square root to both sides that are positives)

$\frac{a+b}{2}\geq \sqrt{ab}$ (Divide both sides by 2)

AM-GM is proved. QED[/hide]
Hopefully, I didn't bore some of you with so much details. Happy proving!
3X.lich

Silverfalcon, your proof is wrong.  If you want to prove AM-GM by contradiction (I can't imagine why) you need to show that $AM<GM$ leads to a contradiction.

Indeed, if $a\not=b$, then $a+b<2\sqrt{ab}$ implies $(\sqrt a-\sqrt b)^2<0$, which is impossible.

Thanks, blahblahblah.

I didn't really do that proof so I just guessed. :P

Prove that:

$1+x+x^2+...x^n = \frac {x^{n+1} - 1}{x-1}$

[hide="Proof2"]

First, use when n=1.

This would give on the leftside:

$1+x^1 = 1+x$

And on the right side:

$\frac {x^1+1-1}{x-1} \Rightarrow \frac {x^2-1}{x-1} \Rightarrow \frac {(x+1)(x-1)}{x-1} \Rightarrow x+1$

So, true for when $n=1$.

Let's assume it works for when the value is $n$ and prove for $n+1$.

$1+x+x^2+...x^n + x^{n+1} = \frac {x^{n+2} - 1}{x-1}\\
\text {For Left Side}\\
\frac {x^{n+1} - 1}{x-1} + x^{n+1} = \frac {x^{n+1} - 1 + (x^{n+1})(x-1)}{x-1}\\
\frac {x^{n+1} - x^{n+1} - 1 + x^{n+2}}{x-1}\\
\frac {x^{n+2} - 1}{x-1}\\
\text {Which is same as right side}$

So, this is proven.[/hide]

I hope I can get some more of these to make this thread more things to look. :)

For this proof keep in mind the generalization.

For $x^n - y^n$ the following equality is true.

$x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$

Thus, we can see that for $y=1$, Silverfalcon's formula for $1 + x + x^2 + \ldots + x^{n-1} + x^n$ is correct.

Here is another important proof technique.

This is called Pigeon Hole Principle, simply PHP.

I'm not sure how to put this in normal word so I'll just give an example problem.

Perhaps someone who is more experienced can introduce in normal words.

John has two million candies. It is identified that the number of flavors of John's candies are no more than 5000. Show that two of John's candies must have same flavor.

[hide="PHP1"]

Since John's candies ---Pigeons--- are more than number of flavors ---Pigeon Holes---, there must be more than one candies per each flavor. So, if I say that all million candies are in different flavor, this implies the two million flavors, which contradicts the first statement of only 5000 flavors. Therefore, two of John's candies must have same flavor.[/hide]

Maybe my explanation is not that good :( but the idea lies in there, hopefully. :D

Colouring: Prove that you cannot cover an 8x8 chessboard with two opposite corners taken off with 31 dominos.

Proof: Colour the squares black and white alternately. There are 32 of one colour and 30 of the other. But each domino must cover a black square and a white square, so 31 dominos cover 31 black squares and 31 white squares. Thus it cannot be done.

[quote="white_horse_king88"]I think it would be more effective with 5001 candies and 5000 flavors to show exactly that point. So, the idea is, there are n + 1 candies and n flavors. There must be at least two candies with the same flavor. Basically it's a proof by contradiction. For example, lets say that each of the n + 1 candies has a unique flavor. That would imply that n + 1  <= n, which can't be true. Thus, the original statement is true.[/quote]

Thankyou very much!! :) I've studied PHP in books but I could never understand how to use it in real problems. Now I do!

[quote="xdl615x"]Posting here to push up this thread so i can see it when i get home. Is that allowed? I hope I don't get a thousand PM's calling me spammer.  :([/quote]

It's okay.. For this thread, I was thinking about reviving anyway to introduce some more proof techniques.

[quote="chess64"]I've studied PHP in books but I could never understand how to use it in real problems. Now I do![/quote]

I'm going to try to post some problems using PHP (Pigeon Hole Principle) soon in this thread.  :) Hopefully this will get more attention to this thread.

Nice.

I'll try to post some more problems but these are I can find now.  :D

Source: Mathematical Circles (Russian Experience)

1. Prove

$\frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots 2n} \leq \frac{1}{\sqrt{2n+1}}$

where $n$ is any natural number.

2. Show that $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1-\frac{1}{n^2}\right) = \frac{n+1}{2n}$

3. $\frac{1}{n+1} + \frac{1}{n+2} + \cdots \frac{1}{2n} > \frac{13}{24}$

If you're above Getting Started or experienced in math induction, please leave these to getting starters.

Just wondering whether proof by induction is better or proof like the Gauss one. i mean if we use induction we don't really know  "why" , it's more like forcing it through.

Another thing is that  how is one supposed to know whether to use induction or if there is a better method by just looking at the question?

[quote="joml88"][quote="towersfreak2006"]Here's another way to show $1+2+3+\ldots+n=\frac{(n)(n+1)}{2}$:

Let $S=1+2+3+\ldots+n$.
Then also, $S=n+n-1+n-2+\dots+1$.

Add the two together:
$2S=(1+n)+(2+n-1)+(3+n-2)+\dots+(n+1)$.
$2S=(n+1)+(n+1)+(n+1)+\ldots+(n+1)$.
Since $S$ had $n$ terms, then $2S$ has $n$ terms of $(n+1)$, so:
$2S=(n+1)(n)$.
$S=1+2+3+\ldots+n=\frac{(n+1)(n)}{2}$.[/quote]

That's the same thing that Moubinool did.

As for the Pigeon Hole Principle, here is the stronger version of it:

Given $n$ holes and $kn+1$ pigeons to place in them, there exists a hole with at least $k+1$ pigeons in it.[/quote]

Just for the sidenote, what joml88 posted is called General Pigeon Hole Principle.

My PHP description covers when $k=1$.

Oh, I didn't notice.  :oops:

I don't know how to use Pigeonhole for white_horse_king88's problem.  :?
Maybe something with the Fibonacci sequence? It seems to be happy: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89

Hmm, look at the equation as the equality of two differences. How many unique differences are there?

[quote="white_horse_king88"]Pigeonhole problem:

Given 16 distinct positive integers not exceeding 100, prove that it is possible to choose 4 distinct integers a, b, c and d such that $a+c = b + d$.

Is this statement still true for 15 distinct positive integers?

[/quote]
[hide]
We can rewrite this as $a-b = d-c$.  If we let any pairing of 2 of the 16 numbers (in the form $x-y$ where $x > y$) have a unique difference, then we have $\frac{15(15-1)}{2} = 105$ unique differences.  This implies that there are more than 100, namely 105, positive integers less than 100 which is not true.  It is, therefore, possible to choose 4 distinct integers a, b, c and d such that $a+c = b+d$ from these 16.

I think another way to prove it would be to show that we have $15*14 = 210$ possibilities for differences (either x-y or y-x) and then that there are only 200 possibilities for differences.

[/hide]


7 times a # is the same as 12 more than 3 times the #.find the #

Call the number $ \alpha$.

$ 7 \alpha \equal{} 12 \plus{} 3 \alpha$

$ 4\alpha \equal{} 12$

$ \alpha \equal{} \boxed{3}$


symmetry
linear algebra
linear algebra solved

Prove that any bilinear inner product $g: L\times L\to K$ where the field $K$ has characteristic different from 2 can uniquely be decomposed into a sum of symmetric and antisymmetric inner products.

the solution is just one line but it is much nicer if you think about this yourself with this hint :

suppose $g=k+l$ with k symmetric and l antisymmetric

then $g(v,w) = k(v,w)+l(v,w)$

and $g(w,v)=k(w,v)+l(w,v)=k(v,w)-l(v,w)$  where in the last equation i used symmetry and antisymmetry of k and l respectively

looking at these two equations and using the fact that the characteristic is not two, should immediately help you find k and l, and thus help you prove decomposition

thank u very much.You are right, it's trivial.Just that when I read the problem I didn't quite studied for the subject :oops:


number theory
prime numbers
number theory unsolved

Does there exist infinite, growing sequence of prime numbers $(q_n)$, such that $q_{n+2}+2q_n=3q_{n+1}$ for all natural numbers $n$.

Assume there does exist such a sequence.

It would mean that we can find integers $\alpha,\beta$ s.t. $q_n=\alpha\cdot 2^n+\beta,\ \forall n\ge 0$ (use the characteristic equation of the recurrence). We easily get a contradiction from here.


inequalities
LaTeX
inequalities proposed

Why didn't you post it in LaTeX?  :?: For some LaTeX help, please visit [url=http://www.mathlinks.ro/LaTeX/AoPS_L_About.php]this site[/url].
Anyway, we usually write inequality, not inequalitie :P

His inequality is the following one:

For $a,b,c,d>0$ satisfying $a+b+c+d=5$ and $abc\ge\frac{3}{2},abd\ge\frac{3}{2},bcd\ge\frac{3}{2},cda\ge\frac{3}{2}$ prove that:

$\frac{7}{4}\le\frac{1}{1+\sqrt{abc}}+\frac{1}{1+\sqrt{abd}}+\frac{1}{1+\sqrt{bcd}}+\frac{1}{1+\sqrt{cda}}\le2$

Yes, that is the inequality.So, try to solve it. ;)

$\frac{7}{4}\le\frac{1}{1+\sqrt{abc}}+\frac{1}{1+\sqrt{abd}}+\frac{1}{1+\sqrt{bcd}}+\frac{1}{1+\sqrt{cda}}\le2$

So, the right estimate is clear since $\frac{1}{1+\sqrt{t}}\le\frac{1}{1+\sqrt{\frac{3}{2}}}$ because $t\ge\frac{3}{2}$. So we have:

$V=\frac{1}{1+\sqrt{abc}}+\frac{1}{1+\sqrt{abd}}+\frac{1}{1+\sqrt{bcd}}+\frac{1}{1+\sqrt{cda}}\le\frac{4}{1+\sqrt{\frac{3}{2}}}$. We want to prove that $V\le2$. It is sufficient to prove that $\frac{4}{1+\sqrt{\frac{3}{2}}}\le2\iff2\le1+\frac{\sqrt{3}}{\sqrt{2}}\iff\sqrt{2}\le\sqrt{3}$ what's OK. Note that we have never have equality. Second note that we didn't need the information about the sum..

I will have a look at the first inequality later..

We need that sum for the left inequality. ;)

Simple...But a hell lot's of work for the first one

I know that is easy.But how did you say, it's lot of work.This problem I've composed for 8 grade at a contest in my district.