<s>[INST] Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$ Let $x,y,z$ be real numbers all greater than $1$, then prove that 
$$\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$$
My Attempt:
I am trying to use $A.M>GM$ but am not able to do it
 [/INST] To prove it, let's subtract LHS from RHS and call it $\Delta$:
\begin{align}
\Delta&=\frac{x-1}{y-1}-\frac{x+1}{y+1}+\frac{y-1}{z-1}-\frac{y+1}{z+1}+\frac{z-1}{x-1}-\frac{z+1}{x+1}\\
&= 2\left(\frac{x-y}{y^2-1}+\frac{y-z}{z^2-1}+\frac{z-x}{x^2-1}\right)\\
\end{align}
Assuming $1<x\leq y\leq z$, we can see:
$$\frac{z-x}{x^2-1}\geq \frac{z-x}{y^2-1}$$
and also since $y-z\leq 0$:
$$\frac{y-z}{z^2-1}\geq \frac{y-z}{y^2-1}$$
Therefore:
$$\Delta\geq \frac{2}{y^2-1} \left(x-y+y-z+z-x\right)=0$$
For the other case where $1<x\leq z\leq y$, you can see:
$$\frac{z-x}{x^2-1}\geq \frac{z-x}{z^2-1}$$
and since $x-y\leq0$:
$$\frac{x-y}{y^2-1}\geq \frac{x-y}{z^2-1}$$
and again results into:
$$\Delta\geq 0$$
which completes the proof.
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