Let $x>0$ then prove or disprove that :
$$x^{x^{x^{x^{x^{x}}}}}\geq e^{\left(\frac{\ln^{2}\left(x+1\right)}{\ln\left(2\right)}-\ln\left(2\right)\right)}\geq \left(x^{2}+1\right)0.5$$
My attempt :
Claim :
Let $0<x\leq 1$ and $0<a<1$ and $f(x)=\ln\left(x^{x^{a^{a^{a^{a}}}}}\right)$ then :
$$f''(x)>0$$
Proof :
$$f''(x)= 2 a^{a^{a^a}} x^{a^{a^{a^{a}}}-2} - x^{a^{a^{a^{a}}}-2} + (a^{a^{a^{a}}} - 1) a^{a^{a^{a}}} x^{a^{a^{a^{a}}}-2} \ln(x)$$
It's straightforward since $$0.5\leq a^{a^{a^{a}}}\leq 1$$
So the function $e^{f(x)}$ is log concave it conducts to the inequality :
$$x^{x^{x^{x^{x^{x}}}}}\geq 2\frac{\left(\frac{x+y}{2}\right)^{2\left(\frac{x+y}{2}\right)^{x^{x^{x^{x}}}}}}{y^{y^{x^{x^{x^{x}}}}}}$$
I'm stuck here...Perhaps we can show that the function $e^{f(x)}$ is strong log convex.
Edit 1
As the function seems to be strong log convex on $x,a\in(0,1]$ we have to show something like :
$$x^{x^{x^{x^{x^{x}}}}}\geq\frac{\left(\frac{\left(x+a\right)}{2}\right)^{2\left(\frac{\left(x+a\right)}{2}\right)^{x^{x^{x^{x}}}}}}{a^{a^{x^{x^{x^{x}}}}}}e^{\left(\frac{1}{8}\cdot m\left(x-a\right)^{2}\right)}\geq^{?}e^{\left(\frac{\ln^{2}\left(x+1\right)}{\ln\left(2\right)}-\ln\left(2\right)\right)}$$
Where $$f''(x)\geq 2m>0$$
Edit 2:
It seems that the function $f(x)$ for $a\in(0,1)$ and $x\in(0,2)$ is log convex .
Edit 3 : Some toughts :
Let $b=x^{x^{x^{x}}}=cst$ and introduce the function :
$$h(y)=\frac{\left(\frac{\left(y+x\right)}{2}\right)^{2\left(\frac{\left(y+x\right)}{2}\right)^{b}}}{y^{y^{b}}}$$
Then we compute the derivative :
$$h'(y)=u(y) (y (x + y)^b (b \ln(\frac{y + x}{2}) + 1) - (y + x) y^b (b \ln(y) + 1))$$
Then if $y+x\geq 1$ the derivative is positive and equal to zero if $y=x$
It's trivial remarking that :
$$(y + x) y^b\geq(y + x) y=(y + x) y\geq y (x + y)^b$$
And $u(y)>0$
Now if $x\leq y $ we have :
$$0.5x^2+0.5\leq 0.5y^2+0.5\leq^{?}h(y)$$
$$t(y)=\frac{\left(\frac{\left(x+y\right)}{2}\right)^{2\left(\frac{\left(x+y\right)}{2}\right)^{x^{x^{x^{x}}}}}}{y^{y^{x^{x^{x^{x}}}}}}-0.5y^{2}-0.5$$
$t(y)$ is a concave function for $x,y\in(0,1)$
Edit 5 :
An interesting inequality is :
$$\frac{\left(\frac{\left(x+a\right)}{2}\right)^{2\left(\frac{\left(x+a\right)}{2}\right)^{a^{a^{a^{a}}}}}}{x^{x^{a^{a^{a^{a}}}}}}\geq\frac{\left(\frac{\left(x+a\right)}{2}\right)^{2\left(\frac{\left(x+a\right)}{2}\right)^{a}}}{x^{x^{a}}}$$
For $1\geq x\geq a>0.5$ wich matchs partially with our problem.
Edit 6:We have the inequality for $x\geq a\geq 0.5$: $$x^{a^{a^{a^{a^{a}}}}}\geq \left(x+x^{a^{\left(1+a\right)}}\right)0.5$$
Question:
How to show the first inequality ?
Any ideas is welcome !
Reference :
About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$
