Prove $\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\leq\frac{x-1}{y-1}+\frac{y-1}{z-1}+\frac{z-1}{x-1}$

Question

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

Answer 1

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.

Answer 2

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!