Prove inequality of functions Prove that for all $ x>0,x \neq 1$:
$$
\frac{x\ln x}{x^2 -1}< \frac{1}{2}
$$
I tried to find when the derivative gets zeros but I did not succeed. I also tried lagrange theorem. Any suggestions? Any help will be appreciated!

Let
$$ f(x)=x\ln x-\frac12(x^2-1). $$
Then
$$ f'(x)=\ln x+1-x. $$
Note that if $x>1$ or $0<x<1$, then $f'(x)<0$. So by the Lagrange Mean-Value Theorem, we have
$$ f(x)-f(1)=f'(c)(x-1), c\text{ is between }x\text{ and }1. $$
Noting that $f(1)=0$, we have that if $x>1$, then
$$ f(x)<0 $$
and if $0<x<1$, then
$$ f(x)>0. $$
Both cases lead to the original inequality.
