How to prove $x^{\ln x} > \frac{x}{2} + \frac{1}{2x} $?

Question

How to prove the following? $$x^{\ln x} > \dfrac{x}{2} + \dfrac{1}{2x} \tag{1} $$ for all $x \in \mathbb{R}^+\setminus \{1\}$?

I could prove $ x^{\ln x} > x $ and $ x^{\ln x} > x/2 $ (in the appropriate regions),
using $x^{\ln x - 1} > a$ for $a=1, 1/2$

but not $(1)$. I could do derivatives, but then the question becomes about comparing derivaties, which looks like a more difficult problem.

Answer 1

Clearly, for $t\ne 0$, we have $$ e^{t^2}=\sum_{n=0}^\infty \frac{t^{2n}}{n!}> \sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}=\cosh t.$$ Now take $t=\ln x$, to obtain the required inequality.

Note:

1. $\cosh t = \dfrac{e^t+e^{-t}}{2}$
2. The inequality turns into equality when $t=0$ which is $x=1$