<s>[INST] Prove $\left(\frac{a-b}{a+b}\right)^{11}+\left(\frac{b-c}{b+c}\right)^{11}+\left(\frac{c-a}{c+a}\right)^{11}\leq 1$ where $a,b,c >0$ I have to prove:
$ \displaystyle \tag*{} \left(\frac{a-b}{a+b}\right)^{11}+\left(\frac{b-c}{b+c}\right)^{11}+\left(\frac{c-a}{c+a}\right)^{11}\leq 1$ where $a,b,c >0$
My approach:
$ \displaystyle \tag*{}x \mapsto a-b \\\\ y\mapsto b-c \\\\ z \mapsto c-a$
We get, $x+y+z=0$ and then the inequality becomes:
$\displaystyle \tag*{} \left(\frac{x}{x+2b}\right)^{11}+\left(\frac{y}{y+2c}\right)^{11}+\left(\frac{z}{z+2a}\right)^{11}\leq 1$
I don't know how to proceed from this; Can we use AM-GM inequality to prove? Any hints would be appreciated, thanks.
 [/INST] You can consider the expression
$$f(a,b,c)=\left(\frac{a-b}{a+b}\right)^{n}+\left(\frac{b-c}{b+c}\right)^{n}+\left(\frac{c-a}{c+a}\right)^{n}\leq 1$$
for any odd $n$. $f(a,b,c)$ is invariant under cyclic permutations. Hence, you only have to consider the cases (1) $a<b<c$ and (2) $a>b>c$. Introducing $x=a/b>0$ and $y=b/c>0$, the expression becomes
$$f(x,y)=\left(\frac{x-1}{x+1}\right)^{n}+\left(\frac{y-1}{y+1}\right)^{n}+\left(\frac{1-xy}{1+xy}\right)^{n}\leq 1 \,.$$
The function $\frac{x-1}{x+1}$ is strictly increasing from $-1$ to $1$. Therefore, in the first case $a<b<c$, $x<1$ and $y<1$, the first two terms are negative and the last one positive and $\leq 1$. Overall it is clear that $f(x,y)\leq 1$. In the second case $a>b>c$, $x>1$ and $y>1$, the first two terms are positive and the last one negative. However, since $xy>y$ it is clear that $$\left(\frac{y-1}{y+1}\right)^{n}-\left(\frac{xy-1}{xy+1}\right)^{n} < 0$$
and the result follows.
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