Show an inequality about digamma function Let $\psi$ be the digamma function, i.e. $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$. Show that
for $x>0$,
\begin{equation}
\label{1}
\psi\left(1+\frac{1}{2x}\right)-\psi\left(\frac{1}{2}+\frac{1}{2x}\right)>x-\frac{1}{2}x^2. \tag{1}
\end{equation}
I have tried the using that $t=\frac{1}{2x}$, then the inequality can be rewritten as
$$ \psi\left(1+t\right)-\psi\left(\frac{1}{2}+t\right)>\frac{1}{2t}-\frac{1}{8t^2}$$
holds for all $t>0$.  Using the wolframalpha I get the asymptotic expansion
$$ \psi\left(1+t\right)-\psi\left(\frac{1}{2}+t\right)=\frac{1}{2t}-\frac{1}{8t^2}+\frac{1}{64t^{4}}+O\left(t^{-5} \right)$$
for $t\to +\infty$.
How can I prove inequality (1)? Can someone help me? Thank you very much!
 A: From equation (25) in https://doi.org/10.2298/AADM130124002N, we have
$$
\log \Gamma (t + 1) = \left( {t + \frac{1}{2}} \right)\log t - t + \frac{1}{2}\log (2\pi ) + \frac{1}{{12t}} + \frac{1}{{\pi t^3 }}\int_0^{ + \infty } {\frac{{t^2 }}{{t^2  + s^2 }}s^2 \log (1 - e^{ - 2\pi s} )ds} 
$$
and
$$
\log \Gamma \!\left( {t + \frac{1}{2}} \right) = t\log t - t + \frac{1}{2}\log (2\pi ) - \frac{1}{{24t}} + \frac{1}{{\pi t^3 }}\int_0^{ + \infty } {\frac{{t^2 }}{{t^2  + s^2 }}s^2 \log (1 + e^{ - 2\pi s} )ds} 
$$
for all $t>0$. Accordingly,
$$
\log \frac{{\Gamma (t + 1)}}{{\Gamma \!\left( {t + \frac{1}{2}} \right)}} = \frac{1}{2}\log t + \frac{1}{{8t}} - \frac{2}{{\pi t^3 }}\int_0^{ + \infty } {\frac{{t^2 }}{{t^2  + s^2 }}s^2 \tanh ^{ - 1} (e^{ - 2\pi s} )ds} 
$$
for all $t>0$. Differentiating both sides gives
$$
\psi (t + 1) - \psi \left( {t + \frac{1}{2}} \right) = \frac{1}{{2t}} - \frac{1}{{8t^2 }} + \frac{2}{{\pi t^2 }}\int_0^{ + \infty } {\frac{{3t^2  + s^2 }}{{(t^2  + s^2 )^2 }}s^2 \tanh ^{ - 1} (e^{ - 2\pi s} )ds}  > \frac{1}{{2t}} - \frac{1}{{8t^2 }}
$$
for all $t>0$.
