The following inequalities are true or false?
$ (1)$
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {8abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a^2\plus{} b^2\plus{} c^2\plus{}ab\plus{}bc\plus{}ca)} \ge 2\]
for all $ a,b,c\in[\sqrt2\minus{}1,1]$, with equality when $ (a,b,c)\sim (1,1,1);(1,1,\sqrt2\minus{}1).$

$ (2)$
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a^2\plus{} b^2\plus{} c^2)} \le 2\]
for all $ a,b,c$ be sidelengths of a triangle, with equality when $ (a,b,c)\sim (1,1,1) or (1,1,0) or (1,1,\sqrt2).$

[quote="dduclam"]The following inequalities are true or false?
$ (1)$
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {8abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca)} \ge 2\]
for all $ a,b,c\in[\sqrt2 \minus{} 1,1]$, with equality when $ (a,b,c)\sim (1,1,1);(1,1,\sqrt2 \minus{} 1).$

$ (2)$
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b} \plus{} \frac {4abc(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(a \plus{} c)(b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)} \le 2\]
for all $ a,b,c$ be sidelengths of [b]an acute[/b] triangle, with equality when $ (a,b,c)\sim (1,1,1) or (1,1,0) or (1,1,\sqrt2).$[/quote]
These inequalities are true. But I think they are not very nice. Here is my proof:

(1) Write the inequality as
$ \sum a(a \plus{} b)(a \plus{} c) \plus{} \frac {8abc(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \ge 2(a \plus{} b)(b \plus{} c)(c \plus{} a),$
or
$ \sum a(a \plus{} b)a \plus{} c) \plus{} 4abc \minus{} 2(a \plus{} b)(b \plus{} c)(c \plus{} a) \ge 4abc\left[ 1 \minus{} \frac {2(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca}\right].$
This is equvalent to
$ \sum a(a \minus{} b)(a \minus{} c) \ge \frac {4abc}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca} \sum (a \minus{} b)(a \minus{} c),$
or
$ \sum a(a \minus{} b)(a \minus{} c)(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \minus{} 3bc) \ge 0.$
Let $ X \equal{} a(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \minus{} 3bc)$ and $ Y,Z$ are similar. Since $ a,b,c \in [\sqrt {2} \minus{} 1,1],$ we have $ X,Y,Z \ge 0.$ Indeed,
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \minus{} 3bc \ge a^2 \plus{} ab \plus{} ac \minus{} bc \ge \left(\sqrt {2} \minus{} 1\right)^2 \plus{} \left(\sqrt {2} \minus{} 1\right) (b \plus{} c) \minus{} bc \ge \left(\sqrt {2} \minus{} 1\right)^2bc \plus{} \left(\sqrt {2} \minus{} 1\right) (b \plus{} c)\sqrt {bc} \minus{} bc \ge \left(\sqrt {2} \minus{} 1\right)^2bc \plus{} 2\left(\sqrt {2} \minus{} 1\right) bc \minus{} bc \equal{} 0.$
Now, let us assume that $ a \ge b \ge c.$ We have
$ Z(c \minus{} a)(c \minus{} b) \ge 0,$ $ a(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} ac \minus{} 3bc) \ge b(a^2 \plus{} b^2 \plus{} c^2 \plus{} bc \plus{} ba \minus{} 3ac).$
Therefore
$ \sum X(a \minus{} b)(a \minus{} c) \ge X(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \equal{} Y(a \minus{} b)(a \minus{} c) \plus{} Y(b \minus{} c)(b \minus{} a) \equal{} Y(a \minus{} b)^2 \ge 0.$

(2) Write the inequality as
$ \sum a(a \plus{} b)(a \plus{} c) \plus{} \frac {4abc(ab \plus{} bc \plus{} ca)}{a^2 \plus{} b^2 \plus{} c^2} \le 2(a \plus{} b)(b \plus{} c)(c \plus{} a),$
or
$ 4abc \left( 1 \minus{} \frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2}\right) \ge \sum a(a \plus{} b)(a \plus{} c) \plus{} 4abc \minus{} 2(a \plus{} b)(b \plus{} c)(c \plus{} a).$
This is equivalent to
$ \frac {4abc}{a^2 \plus{} b^2 \plus{} c^2} \sum (a \minus{} b)(a \minus{} c) \ge \sum a(a \minus{} b)(a \minus{} c),$
or
$ \sum a(a \minus{} b)(a \minus{} c)(4bc \minus{} a^2 \minus{} b^2 \minus{} c^2) \ge 0.$
Because $ 4bc \minus{} a^2 \minus{} b^2 \minus{} c^2 \equal{} (b^2 \plus{} c^2 \minus{} a^2) \minus{} 2(b \minus{} c)^2,$ it can be written as
$ \sum a(b^2 \plus{} c^2 \minus{} a^2)(a \minus{} b)(a \minus{} c) \minus{} 2\sum a(a \minus{} b)(a \minus{} c)(b \minus{} c)^2 \ge 0,$
or
$ \sum a(b^2 \plus{} c^2 \minus{} a^2)(a \minus{} b)(a \minus{} c) \ge 0.$
Without loss of generality, we may assume that $ a \ge b \ge c.$ Then, we have $ a \minus{} c \ge \frac {a}{b}(b \minus{} c)$ and $ a \minus{} c \ge \frac {b}{c}(a \minus{} b)$ (becaue $ b \plus{} c \ge a$). According to these two inequalities, we have
$ a(b^2 \plus{} c^2 \minus{} a^2)(a \minus{} b)(a \minus{} c) \ge \frac {a^2(b^2 \plus{} c^2 \minus{} a^2)(a \minus{} b)(b \minus{} c)}{b},$
and
$ c(a^2 \plus{} b^2 \minus{} c^2)(c \minus{} a)(c \minus{} b) \ge \frac {c^2(a^2 \plus{} b^2 \minus{} c^2)(a \minus{} b)(b \minus{} c)}{b}.$
Therefore
$ \sum a(b^2 \plus{} c^2 \minus{} a^2)(a \minus{} b)(a \minus{} c) \ge \frac {(a \minus{} b)(b \minus{} c)}{b} [a^2(b^2 \plus{} c^2 \minus{} a^2) \plus{} c^2(a^2 \plus{} b^2 \minus{} c^2) \minus{} b^2(c^2 \plus{} a^2 \minus{} b^2)] \equal{} \frac {(a \minus{} b)(b \minus{} c)}{b}(a^2 \plus{} b^2 \minus{} c^2)(b^2 \plus{} c^2 \minus{} a^2) \ge 0.$
The proof is completed.

Remark. (2) does not hold for any triangle, it can only hold when $ a,b,c$ are the side-lengths of an acute triangle. That's the reason why I changed your statement a little.


linear algebra
matrix
linear algebra unsolved

Prove that are not matrix $A$ and $B$ $(n$x$n)$ such $AB-BA=I$

if $A$ and $B$ exist such as $AB-BA=I_{n}$, then if we apply Trace in this equality, we get that $Tr(AB-BA)=Tr(I_{n})$, so $0=n$, which isn't possible. so there is no $A$ and $B$ ... i used the propriety that $Tr(AB)=Tr(BA)$

This does depend on field; the above proof works in any field of characteristic zero (such as the rational, real, or complex numbers). In a field of finite characteristic, such as the integers mod $p$ (with $p|n$), the result can fail.

An example for $n=2$, in the field of integers mod $2$:
$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}, B=\begin{bmatrix}1&0\\1&1\end{bmatrix}$.
$AB=\begin{bmatrix}0&1\\1&1\end{bmatrix}, BA=\begin{bmatrix}1&1\\1&0\end{bmatrix}, AB-BA=AB+BA=I$

[quote]$AB=\begin{bmatrix}0&1\\1&1\end{bmatrix}, BA=\begin{bmatrix}1&1\\1&0\end{bmatrix}, AB-BA=AB+BA=I$[/quote]
you sure this is right?? Forgive me if i'm wrong but $AB-BA \neq AB+BA\neq I$
$\begin{bmatrix}0&1\\1&1\end{bmatrix}+\begin{bmatrix}1&1\\1&0\end{bmatrix}=\begin{bmatrix}1&2\\2&1\end{bmatrix}\neq\begin{bmatrix}1&0\\0&1\end{bmatrix}$
$\begin{bmatrix}0&1\\1&1\end{bmatrix}-\begin{bmatrix}1&1\\1&0\end{bmatrix}=\begin{bmatrix}-1&0\\0&1\end{bmatrix}\neq\begin{bmatrix}1&0\\0&1\end{bmatrix}$

yes but he did not say what field we were working on, so i took the good old one

In the field I was working with, $2=0$. That's the key to my example. Over a general field, my $AB-BA$ is $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$, but $1=-1$ in the field with two elements.


The goodfood catering company finds that competitors cater lunch for a group of 100 people for $\$5$ each. The manager of Goodfood calculates that for each 25 cent discount per lunch, it is possible to sell an additional 10 lunches. If each lunch costs Goodfood $\$2$ to prepare, how many lunches should be prepared to maximize the profit?

Where are you having difficulties? It seems like you havne't even thought about the problem and expect someone to do your homework .


function
geometric series
algebra unsolved
algebra

[quote="ZetaX"]I assume the $n$-th term is $\frac{F_n}{10^n}$.
Then just use Binet's forumla, split the two summands in it to two geometrical sums, done.[/quote]

Can you explain the solution with detail please??? And another way without using that formula????

[quote="puuhikki"]Here $F_n$ denotes the $n$th Fibonacci number.  Binet's formula for Fibonacci numbers says that $n$th Fibonacci number equals
\[ F_n=\frac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2^n\sqrt{5}}.  \][/quote]

All right, and how can I use this to solve the original problem???? I have to prove 1º that the sum converges.....

Just set it into the sum and look what happens! It gets the sum of two converging geometric series (or majorize it by $\sum_{n=1}^\infty \frac{1}{5^n}$).

Another way:
The generating function $f(x)=\sum_{n=0}^\infty F_n x^n$ fulfills $f(x)- x = \sum_{n=2}^\infty F_n x^n= \sum_{n=2}^\infty F_{n-1} x^n + \sum_{n=2}^\infty F_{n-2} x^n = x f(x) + x^2 f(x)$, giving $f(x) = \frac{x}{1-x-x^2}$.
Now set $x= \frac{1}{10}$ and you get $\sum_{n=0}^\infty \frac{F_n}{10^n} = f\left( \frac{1}{10} \right) = \frac{10}{89}$


logarithms
function
algebra unsolved
algebra

Slove this equation: $ x\equal{}2^{\frac{x^2\minus{}1}{3}}$

It is clear that this answer $ x\equal{}0$.
If $ x<0 \ \Rightarrow \left\{\begin{array}{ll} x<0 \\ 2^{\frac{x^2\minus{}1}{3}>0} \end{array}\right.$.
So, $ x>0$

$ x \equal{} 2^{\frac{x^2\minus{}1}{3}} \iff \log_2 x \equal{} \frac{x^2\minus{}1}{3}$.

Let $ f(x) \equal{} \log_2 x \minus{} \frac{x^2\minus{}1}{3}$.
$ f '(x) \equal{} \frac{1}{x \ln 2} \minus{} \frac{2}{3}x$
$ f ''(x) \equal{} \minus{}\frac{1}{x^2 \ln 2} \minus{} \frac{2}{3}$
So one extremum. $ \left(x \equal{} \sqrt{\frac{3}{2 \ln 2}}\right)$
$ \left(\frac{9}{5}<\sqrt{\frac{3}{2 \ln 2}}<\frac{18}{7} \ \ \left( \because 1\minus{}\frac{1}{2}\plus{}\frac{1}{3}\minus{}\frac{1}{4}<\ln 2<1\minus{}\frac{1}{2}\plus{}\frac{1}{3} \right)\right)$
Therefore, a solution is two at most.

If $ x\equal{}1 \Rightarrow f(x)\equal{}0 \iff 1\equal{}2^{\frac{1^2\minus{}1}{3}}\equal{}2^0\equal{}1$.
If $ x\equal{}2 \Rightarrow f(x)\equal{}0 \iff 2\equal{}2^{\frac{2^2\minus{}1}{3}}\equal{}2^1\equal{}2$.

Thus, $ x\equal{}1, 2$

Just observe that RHS is convex, and 1, 2 are solutions.

The convexity follows by composition of convex functions $ f \circ g$ where $ f$ is increasing.


superior algebra
superior algebra solved

Let (an) and (bn) be two sequences with values in R* so that:
sin(a_n+b_n)=2a_n+3b_n for all n natural.Prove that if lim a_n=0 then:
a)lim b_n=0;
b)lim a_n/b_n=-2

An other try :

. since a(n) -> 0, b(n) is bounded.

. Let b(f(n)) a convergent subsequence of (exists by Bolzano-Weierstrass) -> b
Then sinb = 3b and b=0

. So b(n) has only one convergent sub-sequence and is bounded so is convergent and b(n) -> 0.

. sin (a(n)+b(n)) = a(n) + b(n) + O((a(n)+b(n))^3) = 2a(n) + 3b(n) (O = big 'O' notation)
or a(n)/b(n) + 2 = O((a(n)+b(n))^3 / b(n))

. (a(n)+b(n))^3 / b(n) = a(n)^3/b(n) + O(b(n)) so it's enough to prove a(n)^3/b(n) -> 0 and it's enough to prove a(n)/b(n) is bounded.

. |3 + 2a(n)/b(n)| <= |1 + a(n)/b(n)| (|sinx|<=|x|) which is possible only if a(n)/b(n)  [-2..-1]


If the mr and mrs smith plans to organise sports lesson for they children. The number of soccer lesson is 3, the number of baseball lesson is 4 and the number of baskeball lesson is 5. If each child can enter a maximum of 2 lessons. Find the minimum number of children in the smith family.

[hide]answer=6[/hide]

[hide]If it is the minimum number of children, they all need to go to 2 lessons. Since it really doesn't matter which lesson they go to, we add up all the lessons and divide by 2 to get 6.[/hide]

[hide]There are 12 lessons total, and if everyone takes two lessons, that gives the minimum, 6.[/hide]

[hide="Really easy"]There are a total of $12$ lessons. It really doesn't matter which class each child goes to, so the answer is $\frac{12}{2}= 6$. [/hide]

You complicate the question by saying: Child A can't go to basketball lessons, Child B can't go to soccer lessons, etc...


[quote="delegat"]If $ x$, $ y$ and $ z$ are positive real numbers prove the inequality:

$ \frac {xy}{xy \plus{} x \plus{} y} \plus{} \frac {yz}{yz \plus{} y \plus{} z} \plus{} \frac {zx}{zx \plus{} x \plus{} y}\leq\frac {x}{2x \plus{} z} \plus{} \frac {y}{2y \plus{} x} \plus{} \frac {z}{2z \plus{} y}$[/quote]

The above is wrong . Just try $ x\equal{}y\equal{}z\equal{}2$ . The following is nice and true
$ \frac {xy}{xy \plus{} x^2 \plus{} y^2} \plus{} \frac {yz}{yz \plus{} y^2 \plus{} z^2} \plus{} \frac {zx}{zx \plus{} x^2 \plus{} y^2}\leq\frac {x}{2x \plus{} z} \plus{} \frac {y}{2y \plus{} x} \plus{} \frac {z}{2z \plus{} y}$

but please do not discuss it !! (it is from a running contest)

[quote="silouan"][quote="delegat"]If $ x$, $ y$ and $ z$ are positive real numbers prove the inequality:

$ \frac {xy}{xy \plus{} x \plus{} y} \plus{} \frac {yz}{yz \plus{} y \plus{} z} \plus{} \frac {zx}{zx \plus{} x \plus{} y}\leq\frac {x}{2x \plus{} z} \plus{} \frac {y}{2y \plus{} x} \plus{} \frac {z}{2z \plus{} y}$[/quote]

The above is wrong . Just try $ x \equal{} y \equal{} z \equal{} 2$ . The following is nice and true
$ \frac {xy}{xy \plus{} x^2 \plus{} y^2} \plus{} \frac {yz}{yz \plus{} y^2 \plus{} z^2} \plus{} \frac {zx}{zx \plus{} x^2 \plus{} y^2}\leq\frac {x}{2x \plus{} z} \plus{} \frac {y}{2y \plus{} x} \plus{} \frac {z}{2z \plus{} y}$

but please do not discuss it !! (it is from a running contest)[/quote]

I think you maybe made a typo....
Maybe it should be like the following:
$ \frac {xy}{xy \plus{} x^2 \plus{} y^2} \plus{} \frac {yz}{yz \plus{} y^2 \plus{} z^2} \plus{} \frac {zx}{zx \plus{} z^2 \plus{} x^2}\leq\frac {x}{2x \plus{} z} \plus{} \frac {y}{2y \plus{} x} \plus{} \frac {z}{2z \plus{} y}$ :maybe:

I've found the following similar result :

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

May be we can discuss this one .