TITLE: About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$
QUESTION [40 upvotes]: Problem: Let $x > 0$. Prove that
$$x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12.$$

Remark 1: The problem was posted on MSE (now closed).
Remark 2: I have a proof (see below). My proof is not nice.
For example, we need to prove that
$\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7} \le 0$ for all $0 < x < 1$ for which my proof is not nice.
I want to know if there are some nice proofs. Also, I want my proof reviewed for its correctness.
Any comments and solutions are welcome and appreciated.
My proof (sketch):
We split into cases:
i) $x \ge 1$:
Clearly, $x^{x^{x^{x^{x^x}}}}\ge x^x$.
By Bernoulli's inequality, we have
$x^x = (1 + (x - 1))^x \ge 1 + (x - 1)x = x^2 - x + 1 \ge \frac12 x^2 + \frac12$. The inequality is true.
ii) $0 < x < 1$:
It suffices to prove that
$$x^{x^{x^{x^x}}}\ln x \ge \ln \frac{x^2 + 1}{2}$$
or
$$x^{x^{x^{x^x}}} \le \frac{\ln \frac{x^2 + 1}{2}}{\ln x}$$
or
$$x^{x^{x^x}}\ln x \le \ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}$$
or
$$x^{x^{x^x}}\ge \frac{1}{\ln x}\ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}.$$
It suffices to prove that
$$x^{x^{x^x}}\ge \frac{7}{12} \ge \frac{1}{\ln x}\ln \frac{\ln \frac{x^2 + 1}{2}}{\ln x}. \tag{1}$$
First, it is easy to prove that $$x^x \ge \mathrm{e}^{-1/\mathrm{e}}
\ge \frac{1}{\ln x}\ln\frac{\ln\frac{7}{12}}{\ln x}.$$
Thus, the left inequality in (1) is true.
Second, let $f(x) = x^{7/12}\ln x - \ln \frac{x^2 + 1}{2}$. We have
\begin{align*}
 f'(x) &= \frac{7}{12x^{5/12}}
 \left(\ln x + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7}\right)\\
 &\le \frac{7}{12x^{5/12}}
 \left(\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7}\right)\\
 &\le 0 \tag{2}
\end{align*}
where we have used
$\ln x \le \frac{3x^2 - 3}{x^2 + 4x + 1}$ for all $x$ in $(0, 1]$.
Also, $f(1) = 0$. Thus, $f(x) \ge 0$ for all $x$ in $(0, 1)$.
Thus, the right inequality in (1) is true.
Note: For the inequality $\frac{3x^2 - 3}{x^2 + 4x + 1} + \frac{12}{7} - \frac{24x^{17/12}}{7x^2 + 7} \le 0$ for all $0 < x < 1$,
we let $x = y^{12}$ and it suffices to prove that
$11y^{47} + \cdots + 3 \ge 0$ (a polynomial of degree $47$, a long expression) for all $0 < y < 1$.
We are done.

REPLY [3 votes]: Denote
$$f(x,y)= x^{x^{\large y}},\quad p(x)=x^x,\quad \varphi(x)=f(x,p(x)),\quad g(x)=f(x,\varphi(x)),\tag1$$
then
$$\tau_3(x)=x^{p(x)} = f(x,x),\quad\tau_4(x)=x^{\tau_3(x)}
=\varphi(x),\quad \tau_5(x)=x^{\tau_4(x)}=f(x,f(x,x)),$$
and the given inequality can be reformulated as
$$\tau_6(x)=g(x)\ge\dfrac{1+x^2}2.\qquad (x\in\mathbb R_+)\tag2$$
The derivatives of power towers are:
$$\begin{cases}
&p'(x) = p(x)(x\ln x)' = p(x)(1+\ln x) = p(x)\ln(ex),\\[4pt]
&\tau'_3(x)=\tau_3(x)(p(x)\ln x)' = \dfrac1x\,\tau_3(x)p(x)(1+x\ln x\ln(ex)) = \dfrac1x\,\tau_3(x)p(x) q_3(x),\\[4pt]
&\varphi'(x)=\tau'_4(x) = \tau_4(x)(\tau_3(x) \ln x)' =\dfrac1x\,\tau_4(x)\tau_3(x)(1+\ln \tau_3(x) q_3(x)),\\[4pt] 
&\tau'_5(x)=\dfrac1x\,\tau_5(x)\tau_4(x)(1+\ln\tau_4(x)(1+\ln\tau_3(x)q_3(x))),\\[4pt] 
&g'(x)=\tau'_6(x) = \dfrac1x\,\tau_6(x)\tau_5(x)(1+\tau_5(x)(1+\ln\tau_4(x)(1+\ln\tau_3(x)q_3(x)))).\\[4pt]
\end{cases}\tag3$$
Also,
\begin{cases}
f'_y(x,y) = f(x,y) (x^y \ln x)'_y = f(x,y)x^y\ln^2x\\[4pt]
f'_x(x,y) = f(x,y)(y x^{y-1}\ln x+x^{y-1})=f(x,y)x^{y-1}\ln(ex^y).
\tag4\end{cases}
From $(3)$ should

$\;x_2=\operatorname{argmin} p(x)=e^{-1}\approx 0.367879,\quad
\min p(x) = e^{-e^{-1}}>0.692200.\qquad\qquad(5)$

$\;q'_3(x)=(1+x\ln x\ln(ex))' = \ln^2 x + 3\ln x +1,\;$
\begin{cases}
x_{31}=\operatorname{argmax} q_3(x)= \exp \dfrac{-3-\sqrt5}2 \approx 0.072946\\
x_{32}=\operatorname{argmin} q_3(x) = \exp \dfrac{-3+\sqrt5}2 \approx 0.682518\\
q_3(x_{31})=1+(2+\sqrt5) \exp \dfrac{-3-\sqrt5}2 \approx1.309005\\
q_3(x_{32})=1+(2-\sqrt5) \exp \dfrac{-3-\sqrt5}2 \approx0.838879.
\tag6\end{cases}
Therefore, $\;q_3(x)\ge q_3(x_{32})\;$ and $\;\tau'_3(x)>0,\;$ wherein $\;q_3(x)\;$ increases when $\;x\in(0,x_{31})\vee(x_{32},\infty)\;$ and decreases when $\;x\in(x_{31},x_{32}).\;$



Denote
\begin{cases}
q_4(x)=1+q_3(x)\ln\tau_3(x),\\
q_5(x)=1+q_4(x)\ln\tau_4(x),\\
q_6(x)=1+q_5(x)\ln\tau_5(x).\tag7
\end{cases}
If $\;\underline{x\in(0,x_{31}]},\;$ then $\;q_4(x)\le 1+\ln \tau_3(x)\le\ln(ex_{31})<0,\;$
i.e. $\;\varphi(x)\;$ decreases, wherein from $(4)$ should
$$g(x)=f(x,\varphi(x))\ge f(x,\varphi(x_{31})) \ge  g(x_{31})
> 0.685790 > \dfrac{1+x^2}2,$$
and inequality $(2)$ holds.
If $\;x\in(x_{31},x_{2}),\;$ then both $\;q_3(x)\;$ and $\;|\ln\tau_3(x)|\;$ decrease, and $\;q_4(x)\;$ increases.
Taking in account the numerical root $\;x_4\approx 0.274689\!\dots\;$ of $\;q_4(x),\;$ easily to get
$$\varphi(0.274)>0.593237,\quad \varphi'(0.274)>-0.003,$$
$$\varphi(0.275)>0.593236,\quad \varphi'(0.275)<0.0014,$$
$$x_4=\operatorname{argmin} \varphi(x) \in(0.274, 0.275),\tag8$$
$$\beta = \min \varphi(x) > \min(0.5930237-0.001\cdot 0.003,
0.593236-0.001\cdot 0.0014),$$
$$\beta = \min \varphi(x) > 0.593234.\tag9$$

If $\;\underline{x\in(x_{31},x_4)},\;$ then

$\;q_4(x)\in(-1.83125,0)\;$ increases,
$\;\varphi(x)\in(0.593234,0.740040)\;$ decreases,
$\;q_5(x)\in(1,1.551300)\;$ decreases,
$\;|\ln\tau_5|\in(0.7658,1.9375)\;$ decreases.

Therefore, $\;q_6(x)\in(-2.006,0.235)\;$ increases.
Taking in account the numerical root $\;x_6\approx 0.225627\;$ of $\;q_6(x),\;$ easily to get
$$g(0.225)>0.5428243,\quad g'(0.225)>-0.0037,$$
$$g(0.226)>0.5428235,\quad g'(0.226)<0.0022,$$
$$x_6=\operatorname{argmin} g(x) \in(0.225, 0.226),\tag{10}$$
$$g(x) > \min(0.5428243-0.001\cdot 0.0037,
0.5428235-0.001\cdot 0.0022)>0.542820,$$
$$g(x) > \dfrac{1+0.275^2}2>\dfrac{1+x_4^2}2,$$
i.e. the inequality $(2)$ holds.

Note that $\;ex_6^\beta >1,\;$ then from $(4)$ follows increasing of $\;f(x,y)\;$ by $x$ and by $y\;$ in $\;(x_6,\infty).\;$ Since $\;\varphi(x)\;$ increases in $(x_4,x_2)$, then $\;g(x)\;$ also increases in $\;(x_4,x_2).\;$
In this way,
$$g(x_4)> g(0.226) > \dfrac{1+0.291^2}2,$$
$$g(0.291) > 0.5525 > \dfrac{1+0.324^2}2,$$
$$g(0.324) > 0.5625 > \dfrac{1+0.35^2}2.$$
Therefore, inequality $(2)$ holds when $\;\underline{x\in(x_4,0.35)}.\;$
Since $\;\varphi(0.35)>0.6,\;$ then it suffices to prove the inequality
$$h(x)=x^{0.6}\ln x - \ln\dfrac{1+x^2}2 \ge 0.\qquad (\underline{x\in(0.35,\infty)}).\tag{11}$$
Taking in account that $\;h(1)=0\;$ and
$$h'(x)=x^{-0.4}(1+0.6 \ln x)-\dfrac{2x}{x^2+1},$$
it suffices to prove that the function
$$\psi(y) = (y^5+y^{-5})(1+3\ln y)-2y^2$$
is negative when $\;y\in(0.81,1)\;$ and positive when $y>1.$
$$\textbf{Case}\;y\in(0.81,1).$$
$\;\ln y= \ln(1+y-1)=y-1-\frac12(y-1)^2+\frac13(y-1)^3-\dots\le \frac12(y-1)(3-y),\;$
$$\psi(y) \le 2+3(y-1)(3-y)-2y^2=(y-1)(7-5y) < 0.$$
Proved.
$$\textbf{Case}\;y\in(1,\infty).$$
Let $\;z=\ln y > 0,\;$ then
$$\dfrac12\psi\left(e^z\right) = (3z+1)\cosh 5z - e^{2z} = \cosh 5z - \cosh 2z +3z\cosh 5z - \sinh 2z,$$
$$\dfrac12\psi\left(e^z\right) > z\sum\limits_{n=0}^\infty \left(\dfrac{3\cdot25^n}{(2n)!}-\dfrac{2\cdot 4^n}{(2n+1)!}\right)z^{2n} > 0.$$
Proved.
Therefore, inequality $(11)$ holds, and $(2)$ holds too.