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.

If $x > 1$ we prove equivalent inequality: $2x \ln x \leq x^2 - 1 \iff 2x\ln x - x^2 + 1 \leq 0$. 
Look at $f(x) = 2x \ln x - x^2 + 1$ for $x > 1$. We have $f'(x) = 2\ln x + 2 - 2x$, and $f''(x) = \dfrac{2}{x} - 2 < 0$. So $f'(x) < f'(1) = 0$. So $f(x) < f(1) = 0$, and this means $2x\ln x \leq x^2 - 1$.
If $0 < x < 1$ we prove equivalent inequality: $x^2 - 1 \leq 2x\ln x \iff x^2 - 1 - 2x\ln x \leq 0$.
Look at $f(x) = x^2 - 1 - 2x\ln x$ on $0 < x < 1$. $f'(x) = 2x - 2\ln x - 2$, and $f''(x) = 2 - \dfrac{2}{x} < 0$. 
So $f'(x) > f'(1) = 0 \implies f(x) < f(1) = 0 \implies x^2 - 1 \leq 2x\ln x$. Done.
