Conjecture about the minimum of the Gamma's function

Question

Well first a conjecture :

Does $\exists a_k\in[1/4,1/3]$, $0<x\leq 1$ and $n$ a natural number such that :

$$\sum_{k=1}^{n}x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(a_k+x!\right)}}}> 4n/5\tag?$$

The equality case is related to the abscissa of the minimum of the Gamma function .

For example we have graphically :

$$f\left(x\right)=\frac{\left(x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{4}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.5}+x!\right)}}}\right)}{3}>4/5$$

How to (dis)prove it ?

Answer 1

I give here some further terms $a_k$ we have graphically :

$$f\left(x\right)=\frac{\left(x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{4}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.5}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.49}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.46}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.4616}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.46}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.4616}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.4605}+x!\right)}}}+x!^{1+x!^{\left(\frac{1}{2}+x!\right)^{\left(\frac{1}{3.46043955}+x!\right)}}}\right)}{10}>4/5$$

It seems to oscillate around $x\simeq 1/(3+0.4605)$ and perhaps it converges .

We have also :

$$8×10^{-12}<f\left(0.461632144968362\right)-0.8<9×10^{-12}$$

Remark

We can also use other $a_k$ always in the interval $[1/4,1/3]$ and if we choose a value $a_k=1/3.4601$ now with $y=0.4601$ and the intersection with $h(x)=x!$ it's half away from the abscissa of the minimum of $h(x)$ .But it seems really too good to be true...