Prove that $\frac{x \log(x)}{x^2-1} \leq \frac{1}{2}$ for positive $x$, $x \neq 1$. I'd like to prove 

$$\frac{x \,\log(x)}{x^2-1} \leq \frac{1}{2} $$ 

for positive $x$, $x \neq 1$.
I showed that the limit of the function $f(x) = \frac{x \,\text{log}(x)}{x^2-1}$ is zero as $x$ tends to infinity. But not sure what to do next.

By the standard inequality $e^x>1+x$, we have $x-1>\log(1+x-1)=\log(x)$ thus
$$\frac{x\log(x)}{x^2-1}<\frac{x(x-1)}{x^2-1}=\frac{x}{x+1}\le \frac{1}{2} \text{ for } x< 1.$$
For $x>1$, note that the derivative is
$$\frac{(\log(x)+1)(x^2-1)-2x^2\log(x)}{(x^2-1)^2}=\frac{x^2-1-(x^2+1)\log(x)}{(x^2-1)^2}$$
which is negative since the numerator is $0$ at $1$, clearly negative for large $x$, and can only be $0$ when $\log(x)=\frac{x^2-1}{x^2+1}$, but the derivative of $\log(x)$ is $1/x$ while the derivative of the $\frac{x^2-1}{x^2+1}$ is $4/(x^2+1)^2<1/x$ for $x>1$ so this can never be the case. Finally, since the limit as $x\to 1$ is
$$\lim\limits_{x\to 1}\frac{x\log(x)}{x^2-1}=\lim\limits_{x\to 1}\frac{\log(x)+1}{2x}=\frac{1}{2}$$
by L'Hopital's rule, the desired inequality follows.
