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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.

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    $\begingroup$ Just mentioning that this can not be generalized further. Plots with WolframAlpha show that ${^{n}x} \ge (x^2+1)/2$ for $n=2, 4, 6$, but not for $n=8$. $\endgroup$ – Martin R Jun 7 at 7:18
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    $\begingroup$ This is what I see for $n=8$: wolframalpha.com/input/… and the inequality does not hold near $x=0.2$. $\endgroup$ – Martin R Jun 7 at 8:11
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    $\begingroup$ @RiverLi see math.stackexchange.com/questions/3784218/… .Perhaps it could be inspire you...? $\endgroup$ – Erik Satie Jun 7 at 8:25
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    $\begingroup$ @MartinR Yes, you are right. $\endgroup$ – River Li Jun 7 at 8:32
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    $\begingroup$ @Thomas For degree $47$: First let $y = \frac{1}{1 + z}$ for $z \ge 0$, and we need to prove that $F(z) \ge 0$ for all $z \ge 0$ where $F(z)$ is a polynomial of degree $47$ (also long expression). Then we split into two cases: (1) If $z \ge 1/10$, by letting $z = 1/10 + u$ for $u \ge 0$, we have $F(1/10 + u)$ is a polynomial in $u$ with non-negative coefficients. True. (2) If $0 < z < 1/10$, by letting $z = \frac{1}{10}\cdot \frac{1}{1 + v}$ for $v > 0$, we have $(1 + v)^{47}F(\frac{1}{10}\cdot \frac{1}{1 + v})$ is a polynomial in $v$ with non-negative coefficients. True. We are done. $\endgroup$ – River Li Jun 10 at 0:33
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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.

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    $\begingroup$ Nice.(+1) The idea is that for each appropriate interval, use the bound of the form $^6 x \ge x^{x^a}$. The remaining is whether we can prove $x^{x^a} \ge \frac{x^2 + 1}{2}$ easily. For example, for Case 3, it is easy to prove that $h(x) = 2 - x^{0.5877} - x^{-0.4123}$ is concave, so we only need to check $h(1/2)\ge 0$ and $h(1)\ge 0$. $\endgroup$ – River Li Jun 13 at 4:16
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    $\begingroup$ Your idea of using Pade's approximation is something I've never heard of. It's great to know such tool. I did learn many things from your proof. Thank you. $\endgroup$ – Paresseux Nguyen Jun 13 at 4:27
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    $\begingroup$ Thank you for your solution. Your solution is addressed clearly using e.g. ----- End of solution -----------------------. $\endgroup$ – River Li Jun 13 at 4:31
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    $\begingroup$ It seems that $^6 x \ge x^{x^{16/27}} \ge \frac{x^2 + 1}{2}$ for all $0 < x < 1$. The left inequality is equivalent to $^4 x \ge 16/27$. $\endgroup$ – River Li Jun 13 at 5:15
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    $\begingroup$ It would be great if someone can prove that inequality. I'm really curious about it. For now, I think I would have some rest. Your ineq has haunted for some days. You may not believe it, I even saw it in my sleep. $\endgroup$ – Paresseux Nguyen Jun 13 at 5:38
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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.

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First fact

$$ f(x)=\frac{-(W(-\ln(x)))}{(\ln(x))}= x^{x^{·^{·^·}}}$$ for $0.38<x<1$

Second fact

It seems that we have on $x\in(0.38,1)$ :

$$0.5x^{2}+0.5< f(x)< x^{x^{x^{x^{x^x}}}}\quad (I)$$

Proof for the RHS

$$x^{x^{x^{x^{x^x}}}}> ^{8}x >\cdots>f(x)$$

See https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0 .theorem p240-241.The solution is convergent for $e^{-e}<x\leq 1$ and since $e^{-e}<0.38$ wich is coherent .

Proof of the LHS

First Case

For the LHS we can substitute $x=e^y$ and multiplying by $y$.

We get:

$$0.5y(e^{2y}+1)\geq -W(-y)$$

Or :

$$-0.5y(e^{2y}+1)\leq W(-y)$$

Or :

$$0.5y(e^{2y}+1)\exp(-0.5y(e^{2y}+1))\geq y$$

Or :

$$u(y)=(\ln(0.5(e^{2y}+1))+(-0.5y(e^{2y}+1)))\leq 0$$

The derivative is :

$$u'(y)= -0.5 (e^{2 y} + 1) - e^{2 y} y + \frac{2 e^{2 y}}{e^{2 y)} + 1}$$

Lemma $x\in(0,1)$:

$$0.5\left(x-\frac{1}{\left(x\right)}\right)\leq\ln(x)$$

the proof is not hard .

Now starting with the substitution $x=e^y$ and by the lemma we have :

$$-0.5 (x^2 + 1) - x^2 \ln(x) + \frac{2 x^2}{x^2+ 1}\leq -0.5(x^{2}+1)-x^{2}\left(\left(x-\frac{1}{x}\right)0.5\right)+\frac{2x^{2}}{x^{2}+1}$$

We get a polynomial with a root in $x=1$ .Remains to evaluate a cubic polynomial wich is not hard .It show the inequality for $0<x<0.54$ or $\ln(0.38)<y<\ln(0.54)$

Second case

We need to show :

$$\ln(0.5(e^{2y}+1))+(-0.5x(e^{2y}+1))\leq0$$

For that we need a lemma :

Let $-1<y<0$ then we have with $\alpha=\frac{1}{\ln(4)}$:

$$k(x)=\ln((e^{2y}+1))-e^{\left(\alpha\right)2y}\cdot\ln\left(2\right)<0$$

The proof is not hard .

Using this lemma we need to show for $y\in(-0.74,0)$:

$$m(y)=\left(e^{\left(\alpha\right)2y}\cdot\ln\left(2\right)+\ln\left(0.5\right)+(-0.5y(e^{2y}+1))\right)\leq 0$$

$m''(y)$ have only one root expressible in terms of the Lambert's function .We deduce that $m'(y)$ have two roots on $(-0.73,0]$ .Remains to evaluate the function $m(y)$ at $y=-0.73$.

All of this show the first inequality on $(0.38,1)$ wich is a hard part .

Third Case

For the other part and in the spirit of Riverli'proof we have $x\in(0,0.38]$ :

$$x^{x^{x^{x^{x^{x}}}}}> x^{x^{x^{x^{0.69}}}}> x^{x^{\frac{1}{\sqrt{\left(3\right)}}}}> \left(x^{2}+1\right)0.5$$

The LHS is equivalent to $x^x\geq e^{-e^{-1}}>0.69$ The middle inequality is equivalent to :

$$x^{x^{0.69}}\geq \frac{1}{\sqrt{3}}$$

Or :

$$\ln(y)y\geq \frac{0.69}{\sqrt{3}}$$ Where $x^{0.69}=y$

Wich is easy using the Lambert's function .

The Rhs is :

$$x^{x^{\frac{1}{\sqrt{\left(3\right)}}}}> \left(x^{2}+1\right)0.5$$

We have for $x\in(0,0.31)$:

$$x^{x^{\frac{1}{\sqrt{\left(3\right)}}}}>e^{x^2-\ln(2)}> \left(x^{2}+1\right)0.5$$

And for $x\in[0.31,0.38]$:

$$x^{x^{\frac{1}{\sqrt{\left(3\right)}}}}> 2.6^{\left(x^{2}-\frac{\ln\left(2\right)}{\ln\left(2.6\right)}\right)}> \left(x^{2}+1\right)0.5$$

Thes two last inequality are not hard using derivatives .

Bonus inequality :Let $x>0$ then we have :

$$x^{x^{x^{x^{x^{x}}}}}\geq e^{\left(\frac{\ln^{2}\left(x+1\right)}{\ln\left(2\right)}-\ln\left(2\right)\right)}\geq \left(x^{2}+1\right)0.5$$

Hope it helps !

Reference :

R. Arthur Knoebel, “Exponentials Reiterated,” The American Mathematical Monthly, No. 4, Vol. 88 (1981), pp. 235-252, Apr. 1981

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    $\begingroup$ But the difference $f(x) - (x^2+1)/2$ (which we want to show to be non-negative) is not convex, therefore I wonder how this helps. $\endgroup$ – Martin R Jun 7 at 8:43
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    $\begingroup$ What I am asking is how this helps to show that $f(x) - (x^2+1)/2 \ge 0$. $\endgroup$ – Martin R Jun 7 at 9:00
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    $\begingroup$ Can you prove those refinements, or are that conjectures? $\endgroup$ – Martin R Jun 7 at 9:55
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    $\begingroup$ I downvoted the answer because it is a collection of claims, without any proof. $\endgroup$ – Martin R Jun 7 at 12:41
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    $\begingroup$ This is a Q&A site. Answers should answer the question. And if you say “$f$ is convex”, “We have the refinement ...”, “We can also use ...”, “And we have ...” then I would expect that you can prove those claims. Otherwise say “$f$ seems to be convex”, “I assume that ...”, etc. $\endgroup$ – Martin R Jun 7 at 12:47
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Remarks: @Erik Satie considered $^6 x \ge \lim_{n\to \infty} {^n}x = -\frac{W(-\ln x)}{\ln x}$ for $(38/100, 1)$. I gave alternative proof of $-\frac{W(-\ln x)}{\ln x} \ge \frac12 x^2 + \frac12$ for all $x$ in $(38/100, 1)$.


Case $x \in (38/100, 1)$:

According to Theorem in [1] (Page 240), we have $\lim_{n\to \infty} {^n}x = -\frac{W(-\ln x)}{\ln x}$ where $W(\cdot)$ is the principal branch of the Lambert W function. Also, we have $^6 x \ge {^8}x \ge {^{10}}x \ge \cdots$ which results in $^6 x \ge \lim_{n\to \infty} {^n}x = -\frac{W(-\ln x)}{\ln x}$.

Let us prove that $-\frac{W(-\ln x)}{\ln x} \ge \frac12 x^2 + \frac12$ for all $x$ in $(38/100, 1)$.

To this end, with the substitution $x = \mathrm{e}^{-y}$ for $y\in (0, -\ln\frac{38}{100})$, we need to prove that $$\frac{W(y)}{y} \ge \frac12 \mathrm{e}^{-2y} + \frac12$$ or $$W(y) \ge \frac12 y\mathrm{e}^{-2y} + \frac12 y.$$ Since $u \mapsto u\mathrm{e}^u$ is strictly increasing on $(0, \infty)$, it suffices to prove that $$W(y)\mathrm{e}^{W(y)} \ge \left(\frac12 y\mathrm{e}^{-2y} + \frac12 y\right) \mathrm{exp}\left({\frac12 y\mathrm{e}^{-2y} + \frac12 y}\right)$$ that is $$y \ge \left(\frac12 y\mathrm{e}^{-2y} + \frac12 y\right) \mathrm{exp}\left({\frac12 y\mathrm{e}^{-2y} + \frac12 y}\right)$$ where we have used the fact $W(y)\mathrm{e}^{W(y)} = y$ for all $y > 0$.

With the substitution $z = \mathrm{e}^{-2y}$, it suffices to prove that, for all $z$ in $(38^2/100^2, 1)$, $$0 \ge \ln \frac{1 + z}{2} - \frac{1 + z}{4}\ln z.$$ The remaining is smooth.


Reference

[1] R. Arthur Knoebel, “Exponentials Reiterated,” The American Mathematical Monthly, No. 4, Vol. 88 (1981), pp. 235-252, Apr. 1981. https://www.maa.org/programs/maa-awards/writing-awards/exponentials-reiterated-0

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    $\begingroup$ Nicely done ! (+1) $\endgroup$ – Erik Satie Jun 16 at 14:58
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    $\begingroup$ @ErikSatie Thanks! $\endgroup$ – River Li Jun 17 at 0:02
  • $\begingroup$ Perfect answer. $\endgroup$ – haidangel Jun 17 at 6:55
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    $\begingroup$ @haidangel Thanks. $\endgroup$ – River Li Jun 17 at 6:56
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A partial answer


First Fact :

For $x\in(0,1)$ we have :

$$x^{x^{x^{x^{x^{x}}}}}\geq x^{x^{x}}$$

Proof: see the Reference in my other answer .


Second Fact

For $x\in(0,1)$ we have :

$$ x^{x^{x}}\geq x^{\left(1+\left(x-1\right)x\right)}$$

Hint :use Bernoulli's inequality.


Third Fact

For $x\in[0.65,1]$ we have :

$$x^{\left(1+\left(x-1\right)x\right)}\geq b(x)=\left(x\left(1+\left(x-1\right)\cdot\left(\left(x-1\right)x\right)+0.5\left(x-1\right)^{2}\cdot\left(\left(x-1\right)x\right)\cdot\left(\left(\left(x-1\right)x\right)-1\right)\right)\right)$$

Rewrite $x^{\left(1+\left(x-1\right)x\right)}=xx^{\left(\left(x-1\right)x\right)}$ and use the binomial theorem for $p(x)=x^a$ at $x=1$ .We stop the power series at the second order .

Remains to show :

$$0.5 (x - 1)^2 (x^5 - 2 x^4 + 3 x^2 - 1)=b(x)-0.5x^2-0.5\geq0$$

Or

$$(x^5 - 2 x^4 + 3 x^2 - 1)\geq 0$$

Wich is left to the reader and easy using derivatives .


A lemma :

We have for $a,x\in(0,1)$:

$$x^{a^{a^{1.86a\left(1+a\left(a-1\right)\right)}}}\leq x^{a^{a^{a^{a^{a}}}}}\quad\quad(I)$$

Wich is a refinement if $x=a$

The inequality $(I)$ is equivalent to :

$$a^{a^{a}}\leq 1.86a\left(1+a\left(a-1\right)\right)$$

It seems that we have for $a\in(0.03,1)$

$$a^{a^{a}}\leq a^{0.86\left(1+\left(a-1\right)a\right)} \leq 1.87a\left(1+a\left(a-1\right)\right)$$

We start from :

$$a^{a^{a}}\leq a^{0.86\left(1+\left(a-1\right)a\right)}$$

Wich is equivalent to :

$$a^{a}\geq 0.86\left(1+\left(a-1\right)a\right)$$

The function $f(a)=a^{a}$ is convex so we have :

$$f(x)\geq f'(b)(x-b)+f(b)$$

Remains to choose judicious points wich is not hard using a graphic so I let it to the reader . Also see the reference .

Now we start from :

$$a^{0.86\left(1+\left(a-1\right)a\right)} \leq 1.87a\left(1+a\left(a-1\right)\right)$$


A trick is : put $a$ in exponent on both side and the inequality have the form :

$$(1.87u)^v\geq v^{0.86u}$$

The inequality in $u,v$ reminds me the inequality :

Let $a,b>0$ and $k\in(0,1)$ then we have :

$$a(1-k)+bk\geq a^{1-k}b^{k}$$

Using this we have :

$$\left(1.87u\left(x\right)\right)^{-1}\left(v\left(x\right)^{\left(\frac{x}{\left(0.86u\left(x\right)\right)}\right)^{-1}}\left(x\right)+u\left(x\right)\cdot1.87\cdot\left(1-x\right)\right)\geq \left(\left(1.87u\left(x\right)\right)^{-x}\right)x^{\left(0.86u\left(x\right)\right)}$$

Where :

$$u(x)=x\left(1+\left(x-1\right)x\right)$$ and : $$v(x)=x$$ and $x\in(0.03,1)$


The rest is smooth using the lemma 7.1 (p.136 see the first reference for that) .

End lemma


Second lemma

Let $a,x\in(0,1)$ then we have :

$$x^{a^{\frac{7}{12}}}\leq x^{a^{a^{1.87a\left(1+\left(a-1\right)a\right)}}}$$

Proof :

It's equivalent to :

$$a^{1.87a\left(1+\left(a-1\right)a\right)}\geq \frac{7}{12}$$

The function :

$$n(a)=a^{1.87a\left(1+\left(a-1\right)a\right)}$$

Is convex on $(0,1)$ so admits a global minimum on $(0,1)$. The rest is smooth again !


End Second lemma

Remains to show for $x,a\in(0,1)$ and $x\geq a$:

$$0.5a^{2}+0.5\leq x^{a^{\frac{7}{12}}}$$

I pursue it later thanks for advices or comments !

Reference :

Vasile Cirtoaje, "Proofs of three open inequalities with power-exponential functions", The Journal of Nonlinear Sciences and its Applications (2011), Volume: 4, Issue: 2, page 130-137. https://eudml.org/doc/223938

https://www.planetmath.org/convexfunctionslieabovetheirsupportinglines

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  • $\begingroup$ When you use binomial theorem for $x^a$ at $x = 1$, i.e. $$x^a = 1 + a(x - 1) + \frac12 a(a - 1)(x - 1)^2 + \mathrm{high\ order\ terms},$$ why $x^a \ge 1 + a(x - 1) + \frac12 a(a - 1)(x - 1)^2$? Do you need to prove it? $\endgroup$ – River Li Jun 17 at 15:45
  • $\begingroup$ @RiverLi yes we need it for a rigorous proof .Have you a reference ? $\endgroup$ – Erik Satie Jun 17 at 17:11
  • $\begingroup$ @RiverLi What do you think about the last inequality ?Found with Desmos and using log it seems easy no ? $\endgroup$ – Erik Satie Jun 17 at 17:48
  • $\begingroup$ $x^a \ge 1 + a(x - 1) + \frac12 a(a - 1)(x - 1)^2$ is not difficult to prove. Your idea is nice for $[0.65, 1]$. (+1) $\endgroup$ – River Li 2 days ago
  • $\begingroup$ For the last inequality, I hope to see you prove it. $\endgroup$ – River Li 2 days ago

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