About the inequality : $x^{x^{x^{x^{x^x}}}}\geq x^{x^{x^{((e-2)(1+e))x\left(1+\sqrt{x}\left((\sqrt{x})^3-1\right)\right)}}}\geq x^{x^{\frac{16}{27}}}$

Question

This inequality is due to user RiverLi :

Let $0<x\leq 1$ then we have :

$$x^{x^{x^{x^{x^x}}}}\geq x^{x^{\frac{16}{27}}} \geq 0.5x^2+0.5$$

I propose another one wich states :

Let $0<x\leq 1$ then prove or disprove we have :

$$x^{x^{x^{x^{x^x}}}}\geq x^{x^{x^{\left(\left(e-2\right)\left(1+e\right)\right)\left(x+x^{3}-x^{\frac{3}{2}}\right)}}}\geq x^{x^{\frac{16}{27}}} \geq 0.5x^2+0.5$$

Some background :

Some days ago user RiverLi proposed an inequality (see reference) with a really nice and clever proof .

So the inequality :

$$x^{x^{x^{x^{x^x}}}} \geq 0.5x^2+0.5$$ is already proved .

So there many tricks we can use to show the refinement I propose .It seems that each inequality is a steps with two exponents .

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My attempt :

The first LHS is equivalent to :

$$x^{x^{x}}\leq \left(e-2\right)\left(e+1\right)x\left(1+x^{0.5}\left(x^{1.5}-1\right)\right)$$

So we can use the lemma (see the reference [1] p136):

Let $0<a\leq 1$ and $c>0$ then we have :

$$a^c\leq(1−c)^2+ac(2−c)−ac(1−c)\ln(a)$$

But it doesn't works...

So if we substitute $y^2=x$ the function :

$$h(y)=\left(\left(e-2\right)\left(e+1\right)y^{2}\left(1+y\left(y^{3}-1\right)\right)\right)$$

Is convex for $y\in[0,1]$ .So we can use the same argument as below.A good value is $y=0.51$ it gives us for $x\in(0.14,0.9)$:

$$\left(e-2\right)\left(e+1\right)x\left(1+x^{0.5}\left(x^{1.5}-1\right)\right)\ge 1.193075\left(x^{0.5}-0.51\right)+0.387383\ge x^{x^x}$$

So next we can take the log on both side and use the power series of $x^x$ around $x=1$ wich is well-know .Furthermore I build an inequality wich states :

let $x\in[\frac{1}{5},\frac{3}{5}]$ it seems we have :

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

Unfortunately we cannot use the inequality above .Anyway since $f(x)=x^x$ is convex on $(0,1)$ we can use the same way to get someting like $x,x_0\in(0.14,0.9)$:

$$x^{x^x}\leq\exp(\ln(x)(f'(x_0)(x-x_0)+f(x_0))\leq^{?} 1.193075\left(x^{0.5}-0.51\right)+0.387383$$

To works we need to have $x$ near $x_0$ wich is a little bit embarrassing .

Edit :

We introduce the function :

$$t(x)=x^{2x^{x}}$$

It seems that $t(x)$ is convex on $(0,1)$ so we can get inequality like $x\in(0.15,0.19)$ :

$$x^{2x^x}\leq \left(\frac{\left(t\left(0.15\right)-t\left(0.19\right)\right)}{0.15-0.19}\left(x-0.15\right)+t\left(0.15\right)\right)\leq \left(1.193075\left(x^{\left(0.5\right)}-0.51\right)+0.387383\right)^{2}$$

Remains to show the convexity of this function $t(x)$.To show it we stop at the first derivative wich seems to be the product of two increasing function we have :

$$a(x)=x^{2x^{\left(x\right)}+x-1.376}$$

It seems that $a(x)$ is increasing on $(0,1)$

And :

$$b(x)=x^{1.376}(\ln(x))^{2}+\ln(x)x^{1.376}+x^{0.376}$$

It seems that $b(x)$ is increasing on $(0,1)$.

And we have :

$$t'(x)=a(x)* b(x)$$.

Edit 2 : Some ideas to show if $a(x)$ and $b(x)$ are increasing :

For $a(x)$: We use the formula over the reals :

$$(a+b)(a-b)=a^2-b^2$$

Then take the logarithm .

For $b(x)$ differentiate and we can substitute as $x=y^{\frac{1}{1.376}}$ before.

Correct me if I'm wrong

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For the second it's equivalent to :

$$g(x)=\left(x^{\left(e-2\right)\left(e+1\right)x\left(1+x^{0.5}\left(x^{1.5}-1\right)\right)}\right)>\frac{16}{27}$$

It seems that the function $g(x)$ is convex for $x\in(0,1]$.So we can use the fact , convex functions lie above their supporting lines .

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A related result :Let $x\in(0,1)$ then it seems we have :

$$x(x^{2}-2x+2)\geq x^{x^{x}}$$

To show it we can use the convexity on $(0,1)$ of :

$$v(x)=\frac{\ln\left(x(x^{2}-2x+2)\right)}{\ln\left(x\right)}$$

And :

$$j(x)=x^x$$

We can improve the inequality we have on $(0,1)$:

$$\left(x^{1.15}\left(x^{2}-2x+2\right)\right)^{0.85}\geq x^{x^x}$$

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Question :

How to (dis)prove it ?

Reference :

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

[1] : 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

Answer 1

Not really an answer just for the fun :

We show the inequality for $0.25<x\leq e^{-1}$ :

$$x^{x^{x^x}}>\frac{16}{27}\quad (T)$$

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First claim :

The function $f(x)=x^{x^{x^x}}$ is convex on the interval above .

Proof :

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 .

$$R'(x)=\left(x\cdot\left(\ln(x)\right)^{\left(2\right)}+x\ln(x)+1\right)x^{\left(x^{x}+x-1\right)}$$

$R'(x)$ is the product of two decreasing positive functions :

$$p(x)=\left(x\cdot\left(\ln(x)\right)^{\left(2\right)}+x\ln(x)+1\right)$$

And $$k(x)=x^{\left(x^{x}+x-1\right)}$$

$p(x)$ is decreasing on $J=(0.25,e^{-1})$ (left to the reader because not hard).

For $k(x)$ is decreasing on $J=(0.25,e^{-1})$ .To show it take the logarithm and differentiate and use appropriate bounds for $n(x)=x^x$

With the arguments above $R(x)$ is concave on $J=(0.25,e^{-1})$

We deduce that the function $R(x)^{R(x)}$ is convex on $I$ . To show it differentiate twice and use a general form like : $(n(m(x)))''=\left(R(x)^{R(x)}\right)''$ 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 before divided by $x^x$

In the last link all is positive on $J=(0.25,e^{-1})$ taking the function $g(x)=\ln\left(R(x)^{R(x)}\right)$

We conclude that $x^{x^{x^x}}$ is convex on $J=(0.25,e^{-1})$ .

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Now to show $(T)$ remains to choose a judicious point on the convex fonction $f(x)$ and draw a tangent line