Prove that for real numbers $x,y,z>1$,
$$ \frac{x-1}{y-1} + \frac{y-1}{z-1} + \frac{z-1}{x-1} \ge \frac{x+1}{y+1} + \frac{y+1}{z+1} + \frac{z+1}{x+1}$$
It is tempting to use AM-GM for LHS and RHS:
$$ \frac{x-1}{y-1} + \frac{y-1}{z-1} + \frac{z-1}{x-1} \ge 3 $$ $$ \frac{x+1}{y+1} + \frac{y+1}{z+1} + \frac{z+1}{x+1} \ge 3$$
but I don't see this can be continued.
I am looking for another solution than the solution written below:
Solution 1:
Notice
$$\frac{x-1}{y-1} - \frac{x+1}{y+1} = \frac{(y+1)(x-1) - (y-1)(x+1)}{y^{2}-1} = 2\frac{(x-y)}{y^{2}-1}$$ so it is non-negative if $x \ge y$. Similarly for the others.
The problem's inequality is equivalent with
$$ \frac{(x-y)}{y^{2}-1} + \frac{(y-z)}{z^{2}-1} + \frac{(z-x)}{x^{2}-1} \ge 0$$
My take is that, WLOG: let $x = \max \{x,y,z\}$, and if $x \ge y \ge z$:
write
$$ \frac{(x-y)}{y^{2}-1} + \frac{(y-z)}{z^{2}-1} + \frac{(z-x)}{x^{2}-1} $$ $$ = \frac{(x-y)}{y^{2}-1} + \frac{(y-z)}{z^{2}-1} + \frac{(z-x+y-y)}{x^{2}-1} $$ $$ = \frac{(x-y)}{y^{2}-1} + \frac{(y-z)}{z^{2}-1} + \frac{-(y-z) -(x-y)}{x^{2}-1} $$ $$ = (x-y) \frac{x^{2}-y^{2}}{(x^{2}-1)(y^{2}-1)} + (y-z) \frac{y^{2}-z^{2}}{(y^{2}-1)(z^{2}-1)} \ge 0$$
and similarly for $x \ge z \ge y$.