prove an inequality containing digamma function

Question

I wonder how to prove this or if it is correct since that from numerical result, it is correct $$ (2 \log 2-1) x^{2}+(2 \log 2+2) x+2+\left(2 x^{2}+2 x\right) \Psi(x)>0 $$ where $x\in(0,+\infty)$

Answer 1

We split into two cases:

1. $x > 1$:

By using $\Psi(x) \ge \ln x - \frac{1}{x}, \ \forall x > 1$, it is easy to prove the desired inequality.

2. $0 < x \le 1$:

Let $$f(x) = \frac{(2 \log 2-1) x^{2}+(2 \log 2+2) x+2}{ 2 x^{2}+2 x} + \Psi(x).$$ We have \begin{align} f'(x) &= - \frac{3x^2 + 4x + 2}{2(x^2 + x)^2} + \Psi'(x) \\ &= - \frac{3x^2 + 4x + 2}{2(x^2 + x)^2} + \int_0^\infty \frac{t\mathrm{e}^{-xt}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\ &= - \frac{3x^2 + 4x + 2}{2(x^2 + x)^2} + \int_0^\infty t\mathrm{e}^{-xt} \mathrm{d} t + \int_0^\infty \frac{t\mathrm{e}^{-xt}\mathrm{e}^{-t}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\ &= - \frac{1}{2(x+1)^2} + \int_0^\infty \frac{t\mathrm{e}^{-xt}\mathrm{e}^{-t}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\ &> - \frac{1}{2} + \int_0^\infty \frac{t\mathrm{e}^{-t}\mathrm{e}^{-t}}{1 - \mathrm{e}^{-t}} \mathrm{d} t\\ &= - \frac{1}{2} - 1 + \frac{1}{6}\pi^2 \\ &> 0 \end{align} where we have used $\Psi'(x) = \int_0^\infty \frac{t\mathrm{e}^{-xt}}{1 - \mathrm{e}^{-t}} \mathrm{d} t$.

Also, $\lim_{x\to 0^{+}} f(x) > 0$. Thus, $f(x) > 0$ for all $x \in (0, 1]$.

We are done.