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.
 A: Note that for $x>0$ and $t\in[1,x]$ (or $[x,1]$ if you prefer) $$\frac{1}{t}\leq \frac{1-t}{x}+1.$$ The right hand side is a line in the variable $t$ through the points $(1,1)$ and $(x,1/x)$.  Therefore for $x\geq 1$ $$\log(x)=\int_1^x\frac{dt}{t}\leq \int_1^x\left(\frac{1-t}{x}+1\right)dt=\frac{x^2-1}{2x}.$$  For $x\in(0,1]$ the inequality reverses (because integration is in the negative direction now).  Your result follows from these inequalities.
A: 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.
A: 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.
A: 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 $0\lt x\lt1$.
