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.

Given the function
$$
f(x)=\frac{x\log(x)}{x^2-1}\tag{1}
$$
note that
$$
f(1/x)=f(x)\tag{2}
$$
Therefore, we only need to prove the inequality for $x\gt1$.
For $x\gt1$, Consider
$$
g(x)=x\log(x)-\frac{x^2-1}2\tag{3}
$$
we have
$$
g'(x)=1+\log(x)-x\tag{4}
$$
and
$$
g''(x)=\frac1x-1\tag{5}
$$
Thus, we have that $g(1)=g'(1)=0$ and $g''(x)\le0$ for $x\ge1$ Therefore, for $x\ge1$,
$$
g(x)\le0\tag{6}
$$
Equation $(3)$ and inequality $(6)$ yield that for $x\gt1$,
$$
\frac{x\log(x)}{x^2-1}\le\frac12\tag{7}
$$
Equation $(2)$ extends $(7)$ to $0\lt x\lt1$.
