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
 A: Let $z=\min\{x,y,z\}$, $x-1=a$, $y-1=b$ and $z-1=c$.
Thus, $c=\min\{a,b,c\}>0$ and  we need to prove that
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+2}{b+2}+\frac{b+2}{c+2}+\frac{c+2}{a+2}$$ or
$$\frac{a}{b}+\frac{b}{a}-2+\frac{b}{c}+\frac{c}{a}-\frac{b}{a}-1\geq\frac{a+2}{b+2}+\frac{b+2}{a+2}-2+\frac{b+2}{c+2}+\frac{c+2}{a+2}-\frac{b+2}{a+2}-1$$ or
$$\frac{(a-b)^2}{ab}+\frac{(c-a)(c-b)}{ac}\geq\frac{(a-b)^2}{(a+2)(b+2)}+\frac{(c-a)(c-b)}{(a+2)(c+2)},$$
which is obvious.
Another way.
We need to prove that: 
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+2}{b+2}+\frac{b+2}{c+2}+\frac{c+2}{a+2}$$ or
$$\sum_{cyc}(a^3b^2-a^2b^2c+2a^3b+2a^2b^2-4a^2bc+4a^2c-4abc)\geq0,$$
which is true by AM-GM and Rearrangement.
Indeed,
$$\sum_{cyc}a^3b^2=a^2b^2c^2\sum_{cyc}\frac{a}{c^2}\geq a^2b^2c^2\sum_{cyc}\frac{a}{a^2}=\sum_{cyc}a^2b^2c;$$
$$\sum_{cyc}a^3c=abc\sum_{cyc}\frac{a^2}{b}\geq abc\sum_{cyc}\frac{a^2}{a}=\sum_{cyc}a^2bc;$$
$$\sum_{cyc}(2a^2b^2-2a^2bc)=\sum_{cyc}(a^2c^2-2a^2bc+a^2b^2)=\sum_{cyc}a^2(b-c)^2\geq0$$ and
$$\sum_{cyc}a^2c\geq3\sqrt[3]{a^3b^3c^3}=\sum_{cyc}abc.$$
Done again! 
Another way.
Let $k>0$, $\frac{2}{k}=t$ and $x-1=ka$, $y-1=kb$ and $z-1=kc$.
Thus, we need to prove that
$$\sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{ka+2}{kb+2}$$ or
$$\sum_{cyc}\frac{a}{b}\geq\sum_{cyc}\frac{a+t}{b+t}.$$
Now, let $f(t)=\sum\limits_{cyc}\frac{a+t}{b+t}$, where $t>0$.
Since $\lim\limits_{t\rightarrow0^+}f(t)=\sum\limits_{cyc}\frac{a}{b}$, we need to prove that $f$ decreases for $t\rightarrow0^+$. 
Indeed,
$$\lim_{t\rightarrow0^+}f'(t)=\lim_{t\rightarrow0^+}\sum_{cyc}\frac{b-a}{(b+t)^2}=\sum_{cyc}\frac{b-a}{b^2}=$$
$$=\sum_{cyc}\left(\frac{b-a}{b^2}+\frac{1}{b}-\frac{1}{a}\right)=-\sum_{cyc}\frac{(a-b)^2}{b^2a}\leq0.$$
Done!
A: 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.
