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.

  • 3
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @RiverLi see math.stackexchange.com/questions/3784218/… .Perhaps it could be inspire you...? $\endgroup$ – Erik Satie Jun 7 at 8:25
  • 1
    $\begingroup$ @MartinR Yes, you are right. $\endgroup$ – River Li Jun 7 at 8:32
  • 2
    $\begingroup$ @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

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".


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 -----------------------

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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Thank you for your solution. Your solution is addressed clearly using e.g. ----- End of solution -----------------------. $\endgroup$ – River Li Jun 13 at 4:31
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


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 $x\in(0,1)$:


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 :


For that we need a lemma :

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


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

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ Can you prove those refinements, or are that conjectures? $\endgroup$ – Martin R Jun 7 at 9:55
  • 5
    $\begingroup$ I downvoted the answer because it is a collection of claims, without any proof. $\endgroup$ – Martin R Jun 7 at 12:41
  • 9
    $\begingroup$ 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

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.


[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

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

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 :

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


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

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 :


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


  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.