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.
 A: Your inequality is
$$\left(\frac{a-b}{a+b}\right)^{11} + \left(\frac{b-c}{b+c}\right)^{11} + \left(\frac{c-a}{c+a}\right)^{11} \leq 1 \tag{1}\label{eq1A}$$
If any $2$ of the variables, e.g., $a$ and $b$, are equal, the LHS of \eqref{eq1A} becomes $0$. Thus, consider they are all different. Next, for simpler algebra & discussions, let
$$d = \frac{a-b}{a+b}, \; \; e = \frac{b-c}{b+c}, \; \; f = \frac{c-a}{c+a} \tag{2}\label{eq2A}$$
Note that $d, e, f \lt 1$, so $d^{11}, e^{11}, f^{11} \lt 1$. There are $3! = 6$ orderings of $a$, $b$ and $c$ into decreasing sizes. With $d^{11}$, $e^{11}$ and $f^{11}$, note $3$ of the size orderings have $1$ value being positive & $2$ values are negative, while $3$ of the orderings have $1$ value being negative & $2$ values being positive. For the first case, \eqref{eq1A} is obviously true.
For the second case, a first size ordering gives
$$a \gt b \gt c \; \; \to \; \; d \gt 0, \, e \gt 0, \, f \lt 0 \tag{3}\label{eq3A}$$
Since $x \ge y \ge 0 \; \to \; x^{11} \ge y^{11}$, then if $-f \ge d$, we get $(-f)^{11} \ge d^{11} \; \to \; d^{11} + f^{11} \le 0$, so \eqref{eq1A} is true, and similarly if $-f \ge e$. Thus, \eqref{eq1A} can only possibly not be true if both $-f \lt d$ and $-f \lt e$. With $-f \lt d$, we have
$$\begin{equation}\begin{aligned}
\frac{a - c}{c + a} & \lt \frac{a - b}{a + b} \\
(a - c)(a + b) & \lt (a - b)(a + c) \\
a^2 + ab - ac - bc & \lt a^2 + ac - ab - bc \\
2ab & \lt 2ac \\
b & \lt c
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, this contradicts \eqref{eq3A}, so this ordering always satisfies \eqref{eq1A}. A second size ordering is with
$$b \gt c \gt a \; \; \to \; \; d \lt 0, \, e \gt 0, \, f \gt 0 \tag{5}\label{eq5A}$$
Here, $-d \lt e$ gives
$$\begin{equation}\begin{aligned}
\frac{b - a}{a + b} & \lt \frac{b - c}{b + c} \\
(b - a)(b + c) & \lt (b - c)(a + b) \\
b^2 + bc - ab - ac & \lt ab + b^2 - ac - bc \\
2bc & \lt 2ab \\
c & \lt a
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
but this contradicts \eqref{eq5A}, so this size ordering also works. Finally, there's
$$c \gt a \gt b \; \; \to \; \; d \gt 0, \, e \lt 0, \, f \gt 0 \tag{7}\label{eq7A}$$
With $-e \lt d \; \to \; e \gt -d \; \to -d \lt e$, we get the same result as in \eqref{eq6A}, i.e., $c \lt a$. Since this contradicts \eqref{eq7A}, this means \eqref{eq1A} also holds with this third size ordering.
This concludes showing \eqref{eq1A} is always true. Note there is nothing particularly special about the power of $11$ as any odd positive integer exponent would work as well, and also the $\le$ in \eqref{eq1A} can be replaced with just $\lt$.
A: 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.
