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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.

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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 $f(1)=\frac{1}{2}\cdot$

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  • $\begingroup$ so maximum is found using derivatives as well? $\endgroup$ – MathIsTheWayOfLife Aug 4 '16 at 7:50
  • $\begingroup$ @MathIsTheWayOfLife Not directly; to evaluate $f(1)$, a limit must be calculated. $\endgroup$ – heropup Aug 4 '16 at 7:51
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    $\begingroup$ Kind of, you compute the limit with e.g. d'Hospital - using the second derivatives of $x\ln(x)$ and $x^2-1.$ $\endgroup$ – gammatester Aug 4 '16 at 7:52
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    $\begingroup$ It is not a requirement to use L'Hopital's rule; e.g., $$\lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{x}{x+1} \lim_{x \to 1} \frac{\log x - \log 1}{x - 1} = \frac{1}{2} \frac{d}{dx} \left[ \log x \right]_{x = 1} = \frac{1}{2} \left[\frac{1}{x}\right]_{x=1} = \frac{1}{2}.$$ $\endgroup$ – heropup Aug 4 '16 at 7:54
  • $\begingroup$ When $x = 1$, $f'(x)$ is indeterminate, although if we fill in the singularity to make $f$ continuous, then $f'(1) = 0$, not strictly negative. Furthermore, there is insufficient justification for the sign of the numerator of $f'(x)$, since it is not immediately obvious that $x^2 (\log x - 1) + \log x + 1 > 0$ for $ x > 1$: the first term is actually negative until $x \ge e$, so in the interval $1 < x < e$, your claim is not yet formally justified. $\endgroup$ – heropup Aug 4 '16 at 15:28
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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.

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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 $\sinh'(0)=1$.

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  • $\begingroup$ This verifies the inequality for $x\gt1$, which is specified in the question. However, we can go further by noting that for all $x\gt0$ and $x\ne1$, this is equivalent to $\frac{\sinh(t)}t\ge1$ for all $t\ne0$. $\endgroup$ – robjohn Aug 4 '16 at 17:26
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$\boldsymbol{x\gt1}$ (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.

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  • $\begingroup$ Alas, $x > \frac{1}{2}(x+1)$ for $x > 1$; otherwise this would be a nice proof. $\endgroup$ – Unit Aug 4 '16 at 14:28
  • $\begingroup$ @Unit: indeed. Thanks. I have provided a, hopefully correct, proof. $\endgroup$ – robjohn Aug 4 '16 at 14:46
  • $\begingroup$ Would the downvoter care to comment? I fixed the answer in response to Unit's comment. If a problem exists, I will fix it or delete this answer. $\endgroup$ – robjohn Aug 4 '16 at 14:56
  • $\begingroup$ The downvoter was Unit. $\endgroup$ – Unit Aug 4 '16 at 15:21
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Bounding the integral by a trapezoid on the interval $[1,x]$ we get for $x>1$ $$\int_1^x\frac{\mathrm{d}t}t < \tfrac12(1+\frac1x)(x-1) = \frac{x^2-1}{2x}.$$

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