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

$$x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq 1$$

The equality case is $x=0.5$.

To show it I have tried to follow the lemma 7.1 and 7.2 of this paper by Vasile Cirtoaje.The problem is that the resulting expression is awful !

I have tried also Bernoulli's inequality with any effect because it's not sharp enough.

Update 18/12/2020 :

It's a another tried .We can build an approximation like this :

Let $0<\beta<x\leq 0.5$ then we have to determine the constants such that:

$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)$$

We have numerically speaking $\frac{115}{100}<\alpha<\frac{125}{100}$

To reduces the gap I have tried to introduce a linear function : $$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)+ax+b$$

But again it's not enough to work so we may consider a general polynomial like :

$$x^{{\operatorname{W}(2ex)}^{2x}}\leq \frac{1}{2}\operatorname{W}^{\alpha}(2ex)+\sum_{k=0}^{n}a_nx^n$$

Well it's a first step and in the future I shall tried to find the coefficients of this general polynomial .

Update 20/12/2020 :

We can reformulate the problem as :

Let $x,y>0$ such that $ye^y+xe^x=2e$ then we have :

$$\left(\frac{xe^x}{2e}\right)^{(x)^{\frac{xe^x}{e}}}+\left(\frac{ye^y}{2e}\right)^{(y)^{\frac{ye^y}{e}}}\leq 1$$

Where I use the inverse function of the Lambert's function .

Well using the form $f(x)=\left(\frac{xe^x}{2e}\right)^{(x)^{\frac{xe^x}{e}}}=g(x)^{h(x)}$ I can show that the function $f(x)$ is convex on $(0,W(2e))$ so (I have tried) we can use Slater's inequality to find an upper bound .Like this it doesn't work . On the other hand we can use Karamata's inequality but I have not tried !


Well If we use Karamata's I have a strategy :

We have by Karamata's inequality and $0\leq\varepsilon_n'\leq\varepsilon_n<y<x$:

$$f(x)+f(y)\leq f(x+\varepsilon_n)+f(y-\varepsilon_n')$$

With $(y-\varepsilon_n')e^{y-\varepsilon_n'}+(x+\varepsilon_n)e^{x+\varepsilon_n}\geq 2e$

Now we want to repeat the process to get a series of inequalities of the kind :

$$f(x)+f(y)\leq f(x+\varepsilon_n)+f(y-\varepsilon_n')\leq f(x+\varepsilon_{n-1})+f(y-\varepsilon_{n-1}')< 1$$

But it's very complicated.


It doesn't work for all the value but I think we have the inequality $y> 0.5 \geq x$ :

$$p(x)=(1-x^{xe^{x-1}})^2+x^{xe^{x-1}} \frac{xe^{x-1}}{2} (2-x^{xe^{x-1}})-x^{xe^{x-1}} \frac{xe^{x-1}}{2} (1-x^{xe^{x-1}}) \ln\left(\frac{xe^{x-1}}{2}\right)$$ We have : $$f(x)+f(y)\leq p(y)+2^{-\varepsilon}p^{1+\varepsilon}(x)< 1$$

With $0\leq \varepsilon \leq\frac{1}{10}$

Where I use the Lemma 7.2 of the paper above .


The last idea :

Using the majorization theorem :

Let $a\geq b>0$ and $c\geq d >0$ and $n$ a natural number large enough such that :

$$a\geq c$$

And :

$$\left(a\frac{n}{n+1}+c\frac{1}{n+1}\right)\left(b\frac{n}{n+1}+d\frac{1}{n+1}\right)\geq cd$$

Then we have :

$$a+b\geq c+d$$

Proof:It's a direct consequence of the Karamata's inequality.

We have another theorem :

Let $2>x,y>0$ ,$n$ a natural number large enough and $\varepsilon>0 $

If we have :

$$xy<1-\varepsilon $$ $$x+y<2-\varepsilon$$ then we have :

$$\ln\left(\frac{n}{n+1}+x\frac{1}{n+1}\right)+\ln\left(\frac{n}{n+1}+y\frac{1}{n+1}\right)\leq 0$$


Example :

Using the theorem of majorization we have ($x=0.4$):

$$(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}< 1-\operatorname{W}^{1.25}(2ex)0.5$$

And :

$$\left(\frac{1}{4000}x^{{\operatorname{W}(2ex)}^{2x}}+\frac{3999}{4000}\operatorname{W}^{1.25}(2ex)0.5\right)\left(\frac{1}{4000}(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}+\frac{3999}{4000}(1-\operatorname{W}^{1.25}(2ex)0.5)\right)< (1-\operatorname{W}^{1.25}(2ex)0.5)\operatorname{W}^{1.25}(2ex)0.5$$

Dividing both side by the RHS and using the second theorem remarking that :

$$\frac{x^{{\operatorname{W}(2ex)}^{2x}}(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}}{\operatorname{W}^{1.25}(2ex)0.5(1-\operatorname{W}^{1.25}(2ex)0.5)}<1-\varepsilon$$

And :

$$\frac{x^{{\operatorname{W}(2ex)}^{2x}}}{\operatorname{W}^{1.25}(2ex)0.5}+\frac{(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}}{(1-\operatorname{W}^{1.25}(2ex)0.5)}<2-\varepsilon\quad (I)$$

Now I think it's easier because we can take the logarithm and study the behavior of the function .

To prove the $(I)$ we can use the bound :

Let $0<x<\frac{1}{100}$ :

$$e^x<(1+x)^2-x$$

Obviously if we study separately the differents elements of the LHS .

Then to study $(I)$ we have a quite good approximation :

Let $0< x \leq \frac{1}{2}$ then we have :

$${\operatorname{W}(2ex)}^{2x}\geq (2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}$$


In fact we have the following refinement on $(0,0.5]$ :

$$x^{{\operatorname{W}(2ex)}^{2x}}+(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}\leq x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}+ (1-x)^{(2(1-x))^{\frac{915}{1000}\left((1-x)\right)^{\left(\frac{87}{100}\right)}}}\leq 1$$


Remarks : The method using the majorization theorem have two advantages . We need to choose two values of the same order with respect to the values in the LHS . One can be inferior (and the other necessary superior).On the other hand the bound with the exponential ,his accuracy depends of the initial approximation in $(I)$ . Finally if we split in two the LHS in $(I)$ and if for a one we prove a stronger result then the other element is slighlty easier to show .


I build an approximation on $(0,1)$ wich have the form :

$$x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\simeq \left(\left(2^{(2x)^{x^{1.25}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.25}}}\right)^{0.5}\quad (S)$$

You can play with the coefficients $-0.25$ and $1.25$ wich are not the best (make me a comment if you have better please:-))


We can slightly improve $(S)$ in using the logarithm we have on $[0.5,1)$:

$$x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\simeq \left(\left(2^{(2x)^{x^{3}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.2}}}\right)^{0.5}-0.5\ln\left(\left(\left(2^{(2x)^{x^{3}}} \frac{x}{2}\right)^{0.5}0.5^{0.5}*0.5^{{(2 (1-x))}^{x^{-0.2}}}\right)^{0.5}\right)+0.5\ln\left(x^{(2x)^{\frac{915}{1000}\left(x\right)^{\left(\frac{87}{100}\right)}}}\right)\quad (S')$$

We can replace the coefficient $\frac{915}{1000}$ by $\frac{912}{1000}$,and $3$ by $3.5$ and finally $-0.2$ by $-0.19$ and I think it's the same order so we can apply the majorization theorem .Ouf!

Any idea to solve it ?

Thanks

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    $\begingroup$ Does “then we have” mean that you know that the inequality holds, or is it a conjecture? Is there any background to this? Why would it be interesting or useful? $\endgroup$ – Martin R Dec 11 '20 at 12:28
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    $\begingroup$ @MartinR I know it's holds checked numerically . $\endgroup$ – Erik Satie Dec 11 '20 at 16:38
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    $\begingroup$ @ErikSatie For $0 < x \le 1/2$, I guess that we may use a polynomial bound for $(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}}$. $\endgroup$ – River Li Dec 21 '20 at 4:09
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    $\begingroup$ @ErikSatie I guess we may find the bounds ${\operatorname{W}(2ex)}^{2x} \ge P(x)$ and $(1-x)^{{\operatorname{W}(2e(1-x))}^{2(1-x)}} \le Q(x)$ where $P(x)$ and $Q(x)$ are polynomials or rational functions (ratio of two polynomials). Then we go to prove that $P(x)\ln x \le \ln (1 - Q(x))$. Then take derivative. But it will be very complicated. $\endgroup$ – River Li Dec 22 '20 at 10:01
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    $\begingroup$ If you have a differentiable function $f$ defined on $(0,1)$, then $f(x)+f(1-x)$ has a maximum or a minimum at $x=1/2$, since $\left(f\left(x\right)+f\left(1-x\right)\right)^{\prime}=f^{\prime}\left(x\right)-f^{\prime}\left(1-x\right).$ Now assume that $f(x)$ is monotone increasing in $(0,1)$, then $f(1-x)$ is monotone decreasing in $(0,1)$ and so, in this interval, $x=1/2$ is the only extrema. After these observations, for proving your inequality it is enough to show that $f(x)$ is montone increasing in $(0,1)$ and at $x=1/2$ we have a maximum of $f(x)+f(1-x)$. $\endgroup$ – Marco Cantarini Dec 23 '20 at 17:35
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To make the problem more symmetric, let $x=t+\frac 12$ and expand the function as Taylor series around $t=0$.

You will have $$f(t)=1+\sum_{n=1}^p a_n t^{2n}$$ where the $a_n$'s are polynomials of degree $2n$ in $k=\log(2)$ $$a_1=\left\{2,-\frac{13}{4},\frac{1}{2}\right\}$$ $$a_2=\left\{\frac{15}{4},-\frac{1607}{192},\frac{439}{96},-\frac{23}{24},\frac{1}{24}\right\}$$ $$a_3=\left\{\frac{14453}{2880},-\frac{331189}{23040},\frac{142597}{11520},-\frac{7 9}{16},\frac{541}{576},-\frac{11}{160},\frac{1}{720}\right\}$$ $$a_4=\left\{\frac{294983}{53760},-\frac{10787687}{573440},\frac{19112773}{860160}, -\frac{1149103}{92160},\frac{368011}{92160},-\frac{5243}{7680},\frac{15}{2 56},-\frac{43}{20160},\frac{1}{40320}\right\}$$ All these coefficients are negative (this is not the case for $n \geq 5$).

Making the coefficients rational $$g(t)=1-\frac{64 t^2}{5119}-\frac{121 t^4}{738}-\frac{261 t^6}{598}-\frac{182 t^8}{865}+\frac{2309 t^{10}}{1084}+\frac{16024 t^{12}}{1381}+\frac{26942 t^{14}}{613}+O\left(t^{16}\right)$$

Using the above terms, the match is almost perfect for $0\leq t\leq 0.4$ .

Between these bounds, $$\int_0^{0.4}\Big[f(t)-g(t)\big]^2\,dt=1.91\times 10^{-10}$$

What would be amazing is to prove that the minimum value of the function is a bit greater than $0.99$.

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  • $\begingroup$ Well Claude what's the problem for $0.4\leq x\leq 0.5$ ?Thanks ! $\endgroup$ – Erik Satie Dec 27 '20 at 14:20
  • $\begingroup$ @ErikSatie. The problem is for $0.4\leq \color{red}{t}\leq 0.5$ $\endgroup$ – Claude Leibovici Dec 28 '20 at 5:53
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Some thoughts

Let me show how to use bounds for the case $0 < x < \frac{1}{10}$.

Denote $F = W(2\mathrm{e}x)^{2x}$ and $G = W(2\mathrm{e}(1-x))^{2(1-x)}$. We need to prove that $x^F + (1-x)^G \le 1$.

Fact 1: If $u > 0$ and $0 \le v \le 1$, then $u^v \ge \frac{u}{u + v - uv}$.
(Note: By Bernoulli inequality, $(\frac{1}{u})^v=(1+\frac{1}{u}-1)^v\leq 1 + (\frac{1}{u}-1)v = \frac{u + v - uv}{u}$.)

Fact 2: $0 \le 5 - 5F \le 1$ for all $x\in (0, 1/2]$.

Fact 3: $1 \le G < 2$ for all $x\in (0, 1/2]$.

Fact 4: $W(y) \ge \frac{y}{y + 1}$ for all $y\ge 0$.
(Hint: Use $W(y)\mathrm{e}^{W(y)} = y$ for all $y\ge 0$ and that $u \mapsto u\mathrm{e}^u$ is strictly increasing on $(0, \infty)$.)

Fact 5: $F \ge \left(\frac{2\mathrm{e}x}{1 + 2\mathrm{e}x}\right)^{2x}$ for all $x > 0$. (Use Fact 4.)

Fact 6: $G = W(2\mathrm{e}(1-x))^{1 - 2x} W(2\mathrm{e}(1-x)) \ge \frac{W(2\mathrm{e}(1-x))^2}{2x W(2\mathrm{e}(1-x)) + 1 - 2x}$ for all $x \in (0, 1/2]$.
(Hint: Use Fact 1, $u = W(2\mathrm{e}(1-x))$, $v = 1-2x$.)

Fact 7: $W(2\mathrm{e}(1-x)) \ge \frac{48}{35} - \frac{3}{5}x$ for all $x$ in $(0, 1/10)$.

Fact 8: $G \ge \frac{9(16-7x)^2}{-1470x^2+910x+1225}$ for all $x$ in $(0, 1/10)$. (Use Facts 6-7.)

Now, by Facts 1-2, we have $$x^F = \frac{x}{x^{1-F}} = \frac{x}{\sqrt[5]{x}^{5 - 5F} } \le x + (x^{4/5} - x)(5 - 5F).$$ (Note: $u = \sqrt[5]{x}, v = 5-5F$.)

By Facts 1, 3, we have $$(1-x)^G = \frac{(1-x)^2}{(1-x)^{2-G}} \le (1-x)^2 + x(1-x)(2-G).$$ (Note: $u = 1-x, v = 2-G$.)

It suffices to prove that $$ x + (x^{4/5} - x)(5 - 5F) + (1-x)^2 + x(1-x)(2-G) \le 1$$ or $$5(x^{4/5} - x)(1 - F) \le x(1-x)(G-1).$$

By Facts 5, 8, it suffices to prove that $$5(x^{4/5} - x)\left(1 - \left(\frac{2\mathrm{e}x}{1 + 2\mathrm{e}x}\right)^{2x}\right) \le x(1-x)\left(\frac{9(16-7x)^2}{-1470x^2+910x+1225}-1\right).$$

Omitted.

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