Show that the function $f(x)=\left(\sqrt{\frac{e}{2}}\cdot\frac{e}{\pi}\right)-\frac{1}{x}+\left(\frac{e}{2}-\frac{x}{e^{x}}\right)-x$ is negative

Hi I proposed a similar question some days ago and I cannot find the answer .Now the problem :

Let $0<x$ then we have :

$$f(x)=\left(\sqrt{\frac{e}{2}}\cdot\frac{e}{\pi}\right)-\frac{1}{x}+\left(\frac{e}{2}-\frac{x}{e^{x}}\right)-x<0$$

It's not hard to show the concavity of the function $f(x)$ for $x>0$ and then find the maxima at $x=1$

So remains to show that the particular value $f(1)$ is less than zero so negative .

Question :

Can we hope to show it by hand without calculated the extrema with a computer ?

Ps:It seems very challenging .

Solution 1:

Use $$e=\frac{193}{71} \quad \quad\sqrt e=\frac{61}{37} \quad\quad \sqrt 2=\frac{99}{70}\quad\quad \pi=\frac{333}{106}$$ and obtain $$f(1)=-\frac{1505695}{33429263274}=-0.0000450412$$ Use $$e=\frac{1071}{394} \quad \quad\sqrt e=\frac{582}{353} \quad\quad \sqrt 2=\frac{577}{408}\quad\quad \pi=\frac{355}{113}$$ and obtain $$f(1)=-\frac{673314589}{61023141268740}=-0.0000110338$$