About the inequality $x^{x^{x^{x^{x^x}}}} \ge \frac12 x^2 + \frac12$

Solution 1:

Here, I give a full solution to the ineq given by RiverLi.
This solution has the advantage of being purely analytic. By "analytic", I mean, it is not based on any "approximate", "graphic" argument (except numeric calculations). I do try to be as clear as possible, however I do omit some explantions for some points because they're not complicated, just lengthy; in those cases, I add graphs to "justify".

Solution

Before going into details, I restate three simple facts we were able to easily verify by hand, and a lemma that is difficult to show.
$$x^x \ge e^{-1/e} \qquad x^{x^{x^x}}>\left(e^{-1/e}\right)^{e^{1/e}} > \underbrace{0.5877}_{=:a} \text{ and } \quad x^{x^{x^{x^{x^x}}}}> (e^{-1/e})^{1/0.587}>0.5343 $$

Lemma 1: $y \mapsto \frac{ \ln( (y+1)/2)}{\ln(y)}$ is a concave function on $(0.2,1)$.
Graph: Here
Demonstration: At the end.

Now we consider four different possible intervals of values of $x$, namely $[0,0.25)$ , $[0.25, 0.5)$, $[0.5 , 1)$ and $[1, +\infty)$ and prove the ineq in each case.

Case 1: When $x \ge 1$.
As @RiverLi has proven previously, $$x^{x^{x^{x^{x^x}}}} \ge x^x \ge x^2-x+1 \ge \frac{1}{2}(x^2+1)$$ Hence the inequality is true.

Case 2: When $x \in [0, 0.25)$, we have $$^6x> 0.534 >1/2( 0.25^2 +1) \ge \frac{1}{2}(x^2+1)$$ So this case holds.

Case 3: When $x \in [0.5,1)$

Consider the function $f(y)=\ln( e^y+1)$ on $(-\infty,0)$. Because its third derivative $f^{(3)}(y)=\frac{e^y(1-e^y)}{ (1+e^y)^3} $ is positive, we have the following usual inequality $$\frac{f(y)-f(z)}{y-z} \ge f'\left( \frac{y+z}{2}\right)$$ Choose $y=\ln(x^2),z=0$, we imply that: $$\frac{ \ln( (x^2+1)/2)}{\ln(x^2)} \ge \frac{x}{x+1}$$ or (note that $0<x<1$) $$\frac{x^2+1}{2} \le x^{ \frac{2x}{x+1}}$$ Besides $$^6x > x^{x^{0.5877}}= x^{ x^{0.5877}}$$ which implies the sufficiency to show $x^{0.5877} \le \frac{2x}{x+1}$, or $$2 \ge x^{0.5877} +x^{-0.4123}$$ Where the maximum of RHS on $(0.5,1)$ is easy to be analysed, which is in fact achieved at $x=1$, thus the ineq holds. See the graph here

Case 4 $x \in [0.25,0.5]$
As argued in the part 3, $^6x > x^{x^{0.5876}}$( I take $0.5876$ instead of $0.5877$ because it's nicer for later), it suffices to show $$\frac{\ln(x^2+1)-\ln(2)}{\ln(x)} \ge x^{0.5876}$$ on $[0.25, 0.5]$ or $$2\frac{\ln(y+1)-\ln(2)}{\ln(y)} - y^{0.2938} \ge 0$$

on $[0.5 ,\sqrt{0.5} ]\subset [0.5, 0.71]$ Indeed, we will prove the following stronger ineq after using Bernoulli's ineq,

$$2\frac{\ln(y+1)-\ln(2)}{\ln(y)}-\left( 0.6^{p} +p0.6^{p-1}(y-0.6) \right)\ge 0$$ where $p=0.2938$
Now, according to our lemma, the left fraction is concave function, which makes LHS is a summ of a concave and a linear function. Hence $LHS$ is concave. That means LHS attains minium at bord, thus $$LHS \ge \min( LHS_{|y=0.5},LHS_{|y=0.71})=0.007\dots>0$$ Graph here Hence the intial ineq holds for the interval $[0.25,0.5]$. Hence the conclusion. $\square$

Side note: We may not use lemma 1 for the demonstration in case 4 ( just mutiplying both side by $\ln(y)$ then analyze). However, I find this messy and tiresome to check.

----- End of solution -----------------------

Appendix:
Demonstration of lemma 1 I start by demonstrating another lemma
Lemma 2 $f,g$ be differentiable functions on $[a,b]$ such that $g(b)=f(b)=0$, $g(x)> 0,g'(x) < 0$ for all $a<x<b$, then if $x \mapsto \frac{f'(x)}{g'(x)}$ is an decreasing function, so is $\frac{f(x)}{g(x)}$.

Demonstration of lemma 2 Noting $h(x)= \frac{f(x)}{g(x)}$, by Cauchy's MVT, there is a number $c$ lying between $(x,b)$ such that: \begin{align}h'(x)&= \frac{f'(x)g(x)-g'(x)f(x)}{g(x)^2}\\ &=\frac{g'(x)}{g(x)}\left( \frac{f'(x)}{g'(x)}- \frac{f(x)-f(b)}{g(x)-g(b)}\right)=\underbrace{\frac{g'(x)}{g(x)}}_{<0}\underbrace{\left( \frac{f'(x)}{g'(x)}-\frac{f'(c)}{g'(c)}\right) }_{\ge 0} \le 0\end{align} Hence the conclusion.

Back to the demonstration of lemma 1
Note $h(x)=\frac{ \ln(x+1)-\ln(2)}{\ln(x)}$,it suffices to prove $h'(x)$ is decreasing. Now let's study $h$, we have: $$h'(x)= \underbrace{\left( x\ln(x)-(x+1)\ln( \frac{x+1}{2})\right)}_{=:f(x)} \frac{1}{\underbrace{x(x+1)\ln(x)^2}_{=g}}$$ (Check the formula's correctness here)
We see that $f(1)=0$, $f'(x)= \ln(x)-\ln\left( (x+1)/2\right) \le 0$. Hence we have first conclusions that $f(x) \ge 0$, $h$ increasing, and $h\le \lim_{x\rightarrow 1}h(x)=1/2$
The we have $g(1)=0$ and $$g'(x)=\ln(x)(\underbrace{2 + 2 x + \ln(x) + 2 x \ln(x)}_{ >0\text{ if } x>0.2}) \le 0$$ Besides, $$\frac{f'(x)}{g'(x)}=\frac{1-\frac{\ln(x+1)/2}{\ln(x)}}{2 + 2 x + \ln(x) + 2 x \ln(x)} =\frac{1-h(x)}{2 + 2 x + \ln(x) + 2 x \ln(x)}$$ is decreasing because the nominator is decreasing and positive ($h$ is increasing) and the denominator is increasing (by simple calculations or check graph here) Thus based on lemma 2, $h'(x)$ is decreasing. Thus conclusion $\square$

P/s: We can even prove that $y \mapsto \frac{\ln( (y+1)/2)}{\ln(y)}$ is concave on $(0,1)$, not just $( 0.2,1)$, but it is not necessary for our goal.

Solution 2:

Alternative solution:

Case $0 < x < 1$:

It is easy to prove that $x^x \ge \mathrm{e}^{-1/\mathrm{e}}$. Thus, we have $$x^{x^{x^x}} \ge x^{x^{\mathrm{e}^{-1/\mathrm{e}}}}. \tag{1}$$

Also, it is easy to prove that $$\ln y - y\mathrm{e}^{-1/\mathrm{e}}\le \ln \ln \frac{12}{7}, \ \forall y > 0.$$ By letting $y = -\ln x$, we have $$x^{x^{\mathrm{e}^{-1/\mathrm{e}}}} \ge \frac{7}{12}. \tag{2}$$

From (1) and (2), we have $$x^{x^{x^x}} \ge \frac{7}{12}$$ and thus $$x^{x^{x^{x^{x^x}}}} \ge x^{x^{7/12}}.$$

It suffices to prove that $$x^{x^{7/12}} \ge \frac12 x^2 + \frac12$$ or $$x^{7/12}\ln x \ge \ln \frac{x^2 + 1}{2}.$$ 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{24}{7x^2 + 7}\cdot {\frac {1189\,{x}^{2}+574\,x-35}{-35\,{x}^{2}+574\,x+1189}}\right)\\ &= {\frac { 7\left( 1-x \right) \left( 1155\,{x}^{5}-16107\,{x}^{4}-53520 \,{x}^{3}+5232\,{x}^{2}+31629\,x-9861 \right) }{12x^{5/12} \left( {x}^{2}+4\,x+1 \right) \left( 7\,{x}^{2}+7 \right) \left( -35\,{x}^{2}+574\,x+1189 \right) }}\\ &\le 0 \end{align*} where we have used $\ln x \le \frac{3x^2 - 3}{x^2 + 4x + 1}$ for all $x$ in $(0, 1]$, and $x^{17/12} \ge {\frac {1189\,{x}^{2}+574\,x-35}{-35\,{x}^{2}+574\,x+1189}}$ for all $x$ in $(0, 1)$. Also, $f(1) = 0$. Thus, we have $f(x) \ge 0$ for all $x$ in $(0, 1)$.
Note: The bounds come from the Pade approximation. For the former, just take derivative. For the latter, we only need to prove the case when ${\frac {1189\,{x}^{2}+574\,x-35}{-35\,{x}^{2}+574\,x+1189}} > 0$. Let $F(x) = \frac{17}{12}\ln x - \ln {\frac {1189\,{x}^{2}+574\,x-35}{-35\,{x}^{2}+574\,x+1189}} $. We have $$F'(x) = -{\frac { 707455\left( x-1 \right) ^{4}}{12 x \left( 1189 \,{x}^{2}+574\,x-35 \right) \left( -35\,{x}^{2}+574\,x+1189 \right) } } < 0. $$ Also, $F(1) = 0$. The desired result follows.

We are done.