Use of rearrangement inequality

Question

Let (x,y,z) be real numbers, each greater than (1). Prove that $$\dfrac{x+1}{y+1}+\dfrac{y+1}{z+1}+\dfrac{z+1}{x+1} \leq \dfrac{x-1}{y-1}+\dfrac{y-1}{z-1}+\dfrac{z-1}{x-1}$$

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I found a solution here, it goes like this:

$\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{2(x-y)}{y^2-1}$, so it suffices to show that$$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$which is just rearrangement, as $x,y,z$ and $\frac{1}{x^2-1} , \frac{1}{y^2-1} , \frac{1}{z^2-1}$ are oppositely sorted.

Now they claim the inequality $$ \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \leq \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1}$$ is true by rearrangement inequality, assuming $x\geq y\geq z$, then $\frac{1}{x^2-1} \leq \frac{1}{y^2-1} \leq \frac{1}{z^2-1}$. From this we get that $\frac{x}{x^2-1} \geq \frac{x}{y^2-1}$ and $\frac{y}{z^2-1} \geq \frac{y}{y^2-1}$, but how do we know $\frac{z}{x^2-1} \geq \frac{z}{z^2-1}$?

Answer 1

Let $(a_1, a_2, a_3)$ be an increasing rearrangement of $(x, y, z)$ and $$ (b_1, b_2, b_3) = \left(\frac{a_3}{a_3^2-1}, \frac{a_2}{a_2^2-1},\frac{a_1}{a_1^2-1}\right) \, . $$ Then $b_1 \le b_2 \le b_3$ and $$ \frac{x}{y^2-1} + \frac{y}{z^2-1} + \frac{z}{x^2-1} = a_1 b_{\sigma(1)}+a_2 b_{\sigma(2)}+a_3 b_{\sigma(3)} $$ for some permutation $\sigma$ of $\{1, 2, 3\}$. The rearrangement inequality states that this is $$ \ge a_1 b_3 + a_2 b_2 + a_3 b_1 = \frac{x}{x^2-1} + \frac{y}{y^2-1} + \frac{z}{z^2-1} \, . $$

Answer 2

Since the RHS of the inequality is cyclic and not symmetric I dont think you can assume $ x\ge y \ge z$, but we do know that $x, y, z$ and $\frac1{x^2-1},\frac1{y^2-1},\frac1{z^2-1}$ will have the opposite order. The rearrangement inequality states that if $x_1 \le x_2 \dots \le x_n$, and $y_1 \le y_2 \dots\le y_n$, then $$x_1y_n+x_2y_{n-1}+\dots + x_ny_1 \le x_1y_{\sigma(1)}+\dots x_ny_{\sigma(n)}$$ for any permutation $\sigma$. Notice that the LHS here always multiplies the smallest $x$ with the biggest $y$, and this immediately implies that your inequality.