Prove or disprove that the function $f(x)=x^{x^{x^{x}}}$ is convex on $(0,1)$ Let $0<x<1$ and $f(x)=x^{x^{x^{x}}}$ then we have :
Claim :
$$f''(x)\geq 0$$
My attempt as a sketch of partial proof :
We introduce the function ($0<a<1$):
$$g(x)=x^{x^{a^{a}}}$$
Second claim :
$$g''(x)\geq 0$$
We have :
$g''(x)=x^{x^{a^{a}}+a^a-2}(a^{\left(2a\right)}\ln(x)+x^{a^{a}}+2a^{a}x^{a^{a}}\ln(x)-a^{a}\ln(x)+2a^{a}+a^{\left(2a\right)}x^{a^{a}}\ln^{2}(x)-1)$
We are interested by the inequality :
$$(a^{\left(2a\right)}\ln(x)+x^{a^{a}}+2a^{a}x^{a^{a}}\ln(x)-a^{a}\ln(x)+2a^{a}+a^{\left(2a\right)}x^{a^{a}}\ln^{2}(x)-1)\geq 0$$
I'm stuck here .

As noticed by Hans Engler we introduce the function :
$$r(x)=x^{a^a}\ln(x)$$
We have :
$$r''(x)=x^{a^a - 2} ((a^a - 1) a^a \ln(x) + 2 a^a - 1)$$
The conclusion is straightforward the function $\ln(g(x))$ is convex so it implies that $g(x)$ is also convex on $(0,1)$.
Now starting with the second claim and using the Jensen's inequality we have $x,y,a\in(0,1)$:
$$x^{x^{a^{a}}}+y^{y^{a^{a}}}\geq 2\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{a^{a}}}$$
We substitute $a=\frac{x+y}{2}$ we obtain :
$$x^{x^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}+y^{y^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}\geq 2\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}$$
Now the idea is to compare the two quantities :
$$x^{x^{x^{x}}}+y^{y^{y^{y}}}\geq x^{x^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}+y^{y^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}$$
We split in two the problem as :
$$x^{x^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}\leq x^{x^{x^{x}}}$$
And :
$$y^{y^{y^{y}}}\geq y^{y^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}}$$
Unfortunetaly it's not sufficient to show the convexity because intervals are disjoint .

A related result  :
It seems that the function :
$r(x)=x^x\ln(x)=v(x)u(x)$ is increasing on $I=(0.1,e^{-1})$ where $v(x)=x^x$ . For that differentiate twice and  with a general form we have :
$$v''(x)u(x)\leq 0$$
$$v'(x)u'(x)\leq 0$$
$$v(x)u''(x)\leq 0$$
So the derivative is decreasing on this interval $I$ and $r'(e^{-1})>0$
We deduce that $R(x)=e^{r(x)}$ is increasing . Furthermore on $I$ the function $R(x)$ is concave and I have not a proof of it yet .
We deduce that the function $R(x)^{R(x)}$ is convex on $I$ . To show it differentiate twice and use a general form like : $(n(m(x)))''=R(x)^{R(x)}$ and we have on $I$ :
$$n''(m(x))(m'(x))^2\geq 0$$
And :
$$m''(x)n'(m(x))\geq 0$$
Because $x^x$ on $x\in I$ is convex decreasing .
Conlusion :
$$x^{x^{\left(x^{x}+x\right)}}$$ is convex on $I$
The same reasoning works with $x\ln(x)$ wich is convex decreasing on $I$ .
Have a look to the second derivative divided by $x^x$
In the last link all is positive on $J=(0.25,e^{-1})$ taking the function $g(x)=\ln\left(R(x)^{R(x)}\right)$

Question :
How to show the first claim ?Is there a trick here ?
Ps:feel free to use my ideas .
 A: Proof:
First, we have that
(1) Midpoint convex implies rational convex
(2) And Rational convex plus continuous implies convex.
See Midpoint-Convex and Continuous Implies Convex for details on this.
So it suffices to prove the function is mid-point convex.  Let $x\in (0,1)$ and let $y\in (x,1)$ then we want to show that
$$f(\frac{x+y}{2}) \le \frac{f(x)+f(y)}{2}$$
That is, we want to show that
$$\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)^{\left(\frac{x+y}{2}\right)}}} \le \frac{x^{x^{x^{x}}}+y^{y^{y^{y}}}}{2}$$
but this follows from the argument in your original post.
A: Promising method :
As in another topic we can split in two the problem as follow :
We take the log and  stop the derivative at the first order and on $\left(\frac{11}{40},1\right)$ the derivative seems to be the product of two positives increasing function wich I call $a(x)$ and $b(x)$ we have :
$$b(x)=\left[x^{\frac{615-1000}{1000}}(x^{x}(x\ln^3(x)+x\ln^2(x)+\ln(x))+1)\right]$$
And :
$$a(x)=x^{\left(x^{x}-\frac{615}{1000}\right)}$$
And :
$$\frac{d}{dx}\ln\left(x^{x^{x^{x}}}\right)=a\left(x\right)\cdot b\left(x\right)$$
So we can take the log again and again in the general case.Warning to the prohibition of calculus.
Edit :
A good substitution for $b(x)$ is $x=e^y$.Have a look at the first derivative .
It seems we have on $y\in(\ln(0.275),0)$ :
$\frac{d}{dy}b(e^y)\geq g(y)$
Where :
$g(y)=\frac{\left(e^{-\frac{385y}{1000}}\left(1000e^{3y}+e^{3y}(1000(y+1)^{2}+615)y^{3}+e^{3y}(1000(y+1)^{2}+3615)y^{2}+e^{3y}(1000\left(y+1\right)(y+3)-385)y-385e^{3y}\right)\right)}{1000}$
Or $g(y)$ equal to :
$$\frac{\left(e^{-\frac{385y}{1000}}e^{3y}\left(1000y^{5}+3000y^{4}+4615y^{3}+8615y^{2}+2615y+615\right)\right)}{1000}$$
Edit : Some ideas to show that $a(x)$ is increasing on $(0,1)$
The third derivative of $a(x)$ seems to be positive so the derivative of $a(x)$ admits a minimum .So the derivative seems to be convex as the sum of two convex function we have :
$$\left(\frac{\left(x^{x}-0.615\right)}{x}\right)''>0$$
And
$$\left(x^{x}\left(\ln^{2}\left(x\right)+\ln\left(x\right)\right)\right)''>0$$
An easier way is to substitute $x=e^y$ and then differentiate .
In the derivative the embarrassing part is :
$$j(y)=e^{\left(\left(e^{y}+1\right)\cdot y\right)}\left(y\left(y+1\right)\right)+e^{y\cdot e^{y}}-0.615$$
It seems we have on $y\in(-1,0)$ :
$$j(y)\geq e^{\left(\left(y+2\right)\cdot y\right)}\left(y\left(y+1\right)\right)+e^{\left(ye^{y}\right)}-0.615\geq 0$$
Where we use the very famous inequality and $x$ a real number:
$$e^x\geq x+1$$
Again it seems that the functions :
$t(y)=e^{\left(\left(y+2\right)\cdot y\right)}\left(y\left(y+1\right)\right)$
And
$k(y)=e^{\left(ye^{y}\right)}$
Are convex for $y\in(-1,0)$ .So we can use the tangent line .
