Help proving an inequality in calculus $0 \leq \frac{x\ln(x)}{x^2-1}\leq \frac{1}{2}$ So as the title states, I need help with proving this inequality:
$$0 \leq \frac{x\ln(x)}{x^2-1}\leq \frac{1}{2} $$
Also an important thisn is that $x>1$.
So I thought I could look different kinds fo x, but it's a subject of calculus, so perhaps it's related to Lagrange's or Rolle's theorem, but i cannot get an idea, i tried changing the inequality but it's not really working, so any help with solution would be really appreciated.
Thank you in advance.
 A: We may write the inequality as $$0 \le \log x \le \frac{x-x^{-1}}{2}.$$  The LHS is trivial.  At $x = 1$, we have RHS equality, so to show that $(x-x^{-1})/2 - \log x \ge 0$, it suffices to show that $$\frac{d}{dx} \left[\frac{x-x^{-1}}{2} - \log x\right] = \frac{(x-1)^2}{2x^2}\ge 0 $$ for $x > 1$.  This of course is also trivial.
A: The function  $f(x) = \cfrac{x\ln(x)}{x^2-1}$   is positive and  monotone decreasing for $x\ge 1$ because 
$$f'(x) = -\frac{x^2(\ln(x)-1)+\ln(x)+1}{(x^2-1)^2} < 0$$
for $x > 1$. The maximium of f(x) is therefore $f(1)=\frac{1}{2}\cdot$
A: By setting $x=e^t$, your inequality is equivalent to:
$$\forall t>0,\qquad 0\leq t \leq \sinh(t)\tag{1} $$
that is trivial since $\sinh(t)$ is a positive convex function on $\mathbb{R}^+$ and $\sinh'(0)=1$.
A: $\boldsymbol{x\gt1}$ (as specified)
Consider
$$
f(x)=2x\log(x)-\left(x^2-1\right)\tag{1}
$$
Since $e^x\ge1+x$ for all $x$, we have $x\ge1+\log(x)$ for all $x\gt0$. Therefore
$$
f'(x)=2+2\log(x)-2x\le0\tag{2}
$$
and $f$ is monotonically decreasing for $x\gt0$. Since $f(1)=0$, we have that $f(x)\le0$ for $x\ge1$.
That is, for $x\gt1$,
$$
2x\log(x)\le\left(x^2-1\right)\tag{3}
$$
Dividing by $x^2-1$, which is positive, and noting that $x\log(x)$ is also positive, gives
$$
0\le\frac{x\log(x)}{x^2-1}\le\frac12\tag{4}
$$

$\boldsymbol{0\lt x\lt1}$ (as a bonus)
For $0\lt x\lt1$, both $x\log(x)\lt0$ and $x^2-1\lt0$. Therefore, since $f$ is decreasing and $f(1)=0$, we have that for $0\lt x\lt1$,
$$
2x\log(x)\ge\left(x^2-1\right)\tag{5}
$$
Dividing by $x^2-1$ now gives
$$
0\le\frac{x\log(x)}{x^2-1}\le\frac12\tag{6}
$$
for $0\lt x\lt1$ as well.
A: Bounding the integral by a trapezoid on the interval $[1,x]$ we get for $x>1$
$$\int_1^x\frac{\mathrm{d}t}t < \tfrac12(1+\frac1x)(x-1) = \frac{x^2-1}{2x}.$$
