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Inspired by this question.

Note that $a,b \in [0,1]$.

In my attempts to crack the link above, I decided to try and generalize and use an inductive method. Having plugged this into Mathematica for $n = 0,1,...,9,10$, I'm fairly confident that this claim is true. I've sunk about two weeks into this and gotten nowhere so I really want to see a proof for this. It's becoming sort of an internal demon of mine.

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Teddy Mishura
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3 Answers3

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This is only a comment, but too long. Here an example for a short consideration, $n\ge 1$:

If we look at $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}\enspace$ and $\enspace\displaystyle \frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ then we see numerically $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}>\frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ for $\enspace 0<x<0.5\enspace$ and because of the symmetrie $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}<\frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ for $\enspace 0.5<x<1$ . This is equivalent to your question.

We can find out, that the maximum of $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}\enspace$ is at $\enspace x_0\enspace$ (depending on $n$ ;

the function is increasing for $\enspace 0<x<x_0$) $\enspace$ and that the maximum of $\enspace\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}\enspace$

is at $\enspace 1-x_0\enspace$ (the function is increasing for $\enspace 0<x<1-x_0$) $\enspace$ with $\enspace x_0<0.5<1-x_0\enspace$

$($e.g. $n:=2$ : $x_0\approx 0,472376195328360)\enspace$ which is interesting,

because at $\enspace x=0.5\enspace$ we have $\displaystyle \frac{x^{(2(1-x))^n}-0.5}{x-0.5}=\frac{0.5-(1-x)^{(2x)^n}}{x-0.5}\enspace$ for all $\enspace n\enspace$

and we’ve got a piece of circumstantial evidence that the inequalities are valid.

---

Numerically we see that $\enspace\displaystyle x^{(2(1-x))^n}+(1-x)^{(2x)^n}\enspace$ has one minimum for $\enspace 0<x<0.5\enspace$ and the maximum is $1$ at $x=0$ and $x=0.5$ .

We have $\enspace\displaystyle\min(x^{(2(1-x))^{n_2}}+(1-x)^{(2x)^{n_2}})<\min(x^{(2(1-x))^{n_1}}+(1-x)^{(2x)^{n_1}})\enspace$ for $\enspace n_1<n_2$ which is a good hint that a proof for $n=1$ can include the proof for $n>1$ .

This special case $n=1$ can be found as $\,$ Proposition 5.2 $\,$ in the article of https://eudml.org/doc/223938 .

user90369
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  • Why not considering function f(x,y)=x^y+y^x. If the derivative of f(x,y) related to x or y is equaled to zero then x or y=0. putting this in the function we find maximum of function to be 1.This posible only when x+y=1. – sirous Apr 06 '17 at 21:12
  • @Sirous : It's not clear what your comment has to do with the problem of the OP. Please: If you have an idea for a proof write it down so everyone can discuss with you. :-) – user90369 Apr 07 '17 at 13:20
  • I mean finding the maximum of function f(x,y)=x^y+y^x by finding its deravative (with respect to x or y) and equal it to zero to find the root and putting this root of function gives the maximum value of f(x,y). Since a abd b have functional relation a+b=1 then relation a^(2b)^n+b^(2a)^n=q has a maximum value. – sirous Apr 07 '17 at 15:39
  • @Sirous : The problem is, that the discussion of the derivatives is more complicate than the original problem. Please write down your answer if you think that your idea works better than in http://www.emis.de/journals/TJNSA/article/Vol%204%20no%202/Vol4_No2_130-137_PROOFS%20OF%20THREE%20OPEN%20INEQUALITIES%20WITH%20....pdf (page 134, Conjecture 5.1) . :-) – user90369 Apr 08 '17 at 21:27
  • Consider f(x,y) =x^2+y^2 with condition x^2+y^2=r^2; f’(x,y)=d(f(x,y))/dx=-2x/2y=0; → x=0, →y^2=r^2 or y=±r. y=r is maximum value of f(x,y) for x<r and y<r . It is also minimum value of the function f(x,y) if x≥r and y≥r. Now we use this method for question: f(a,b) =a^(2b)^n+b^(2a)^n and condition is a+b=1;We take derivative for example on a, we have: f’(a,b)=(2^n).n.a^(n-1).b^(2n).lnb +(2b)^n . a^((2b)^n -1). F’(a,b)=0 → a=0 →f(a,b) = 0 + b^0 =1, which is the maximum value of f(a,b) if a and b<1 and a≠b . It is also minimum of f(a,b) if a and b>1. – sirous Apr 11 '17 at 11:59
  • @Sirous : With your function $f(a,b)$ ($a$ and $b$ independend) a minimum is given by $f_a(a,b)=0$ , $f_b(a,b)=0$ , $f_{aa}(a,b)f_{bb}(a,b)>f_{ab}^2(a,b)$ , $f_{aa}(a,b)>0$ .To prove this analytically (and not numerically) is much more difficult than the original question. --- Once again: IF you think you have an idea for a solution, then write it as an answer, the OP is still waiting for a solution! – user90369 Apr 11 '17 at 13:23
  • a and b are not idependent and have a relation a+b=1.That was my proposal for solution. You say it makes it more difficult. – sirous Apr 11 '17 at 14:47
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    @Sirous : You have used $a$ and $b$ independend for the derivation of $f(a,b)$ and therefore my comment. --- Please don't write so many comments it's much better to answer the question for the OP (where you can write all your ideas). – user90369 Apr 11 '17 at 15:46
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This is a WRONG proof.

Here is an attempt/sketch, definitely not a proof as there is one problematic spot:

$\cos^{2}(x) + \sin^{2}(x) = 1$, raise by $2^n$

$(\cos^{2}(x) + \sin^{2}(x))^{2^n} \ge \cos^{2}(x)^{2^n} + \sin^{2}(x)^{2^n}$

This is true since $n$ is a natural number and we are removing non-negative quantities from the left. We note that the right side contains two quantities which are less then $1$, yet by the inequality we still have:

$\cos^{2}(x)^{2^n} + \sin^{2}(x)^{2^n} \le 1$, now we create two more inequalities by raising by $\cos^{2n}(x)$ and by $\sin^{2n}(x)$

(The following two inequalities I have no proof for): By similar reasoning, we get $\cos^{2}(x)^{2^n\cos^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \le 1$, and

$\cos^{2}(x)^{2^n\sin^{2n}(x)} + \sin^{2}(x)^{2^n\sin^{2n}(x)} \le 1$

If we add these two inequalities and rearrange them we get

$\cos^{2}(x)^{2^n\sin^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \le 2 - (\cos^{2}(x)^{2^n\cos^{2n}(x)} + \sin^{2}(x)^{2^n\sin^{2n}(x)} )$

We note that the functions on the right are of the form $f(x)=x^{2^nx^{2n}}=x^{(2x^2)^n}$ with symmetry $x \to 1-x$. By looking at the derivative (or at the graph) we see that the function in the interval $[0;1]$ attains a minimum in the interval $[\frac{1}{\sqrt e};1]$, but in the interval $[0;1/2 + \epsilon]$ approaches 1 as $n \to \infty$. In other words $f(x) + f(1-x) \ge 1, x \in [0,1]$ by inspection.

Hence: $\cos^{2}(x)^{2^n\sin^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \le 1$

Now set $ a = \cos^2(x), b=\sin^2(x)$ to get:

$(a)^{2^nb^{n}} + (b)^{2^na^{n}} \le 1$ or

$a^{(2b)^n} + b^{(2a)^n} \le 1$ as desired.}

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    I am really sorry, but your sketch is still not useful, because e.g. to use explizit (and not only to see) the derivations are bringing more difficult problems. Only for looking at the plots the function $x^{(2(1-x))^n}+(1-x)^{(2x)^n}$ is enough - we see that it is less then $1$. In your case: to use $\sin$ and $\cos$ maybe has impressed the OP but make no sense, because to use $a$ instead of $\cos^2$ and $b$ instead of $\sin^2$ doesn't change anything in your argumentation. With other words: What is the new idea in your text, which brings us nearer to a solution ? – user90369 Apr 01 '17 at 11:09
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    Let $n=1$ and $x=\frac{\pi}{3}$. Then \begin{align} &\cos^{2}(x)^{2^n\cos^{2n}(x)} + \sin^{2}(x)^{2^n\cos^{2n}(x)} \ &= \cos(x)^{4 \cos ^2(x)}+\sin(x)^{4 \cos ^2(x)} \ &= \frac{1}{2} \left(1+\sqrt{3}\right) > 1. \end{align}. – cafaxo Apr 01 '17 at 11:14
  • Yeah, I should not be doing mathematics very late at night. Can any of the admins delete this answer? @cafaxo absolutely true. –  Apr 01 '17 at 11:42
  • Maybe you will have better ideas later - simply change your answer, e.g. a sketch of a traditional curve discussion. :-) – user90369 Apr 01 '17 at 17:53
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    I restarted the bounty to appease the Gods of Mathematics –  Apr 02 '17 at 19:02
  • @EdvinOrlov: I'm not aware of any God of Mathematics, but it's very nice of you to offer a bounty. I hope a good proof will be given soon (but I don't have any ideas)! =) – user21820 Apr 06 '17 at 17:48
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Claim

Let $0.5\leq x<1$ and $n\geq 2$ a natural number then we have : $$(1-x)^{(2x)^n}+x^{(2(1-x))^n}\leq 1\quad (I)$$

Sketch\Partial (of) proof .

We use a form of the Young's inequality or weighted Am-Gm :

Let $a,b>0$ and $0<v<1$ then we have :

$$av+b(1-v)\geq a^vb^{1-v}$$

Taking account of this theorem and putting :

$a=x^{(2(1-x))^{n-1}}$$\quad$$b=1$$\quad$$v=2(1-x)$ we get $0.5\leq x<1$:

$$x^{(2(1-x))^n}\leq x^{(2(1-x))^{n-1}}2(1-x)+1-2(1-x)$$

Now the idea is to show :

Let $$(1-x)^{(2x)^n}\leq 1-\Big(x^{(2(1-x))^{n-1}}2(1-x)+1-2(1-x)\Big)$$

Or :

$$(1-x)^{(2x)^n}\leq2(1-x)(1-x^{(2(1-x))^{n-1}})$$

Or: $$(1-x)^{(2x)^n-1}+2x^{(2(1-x))^{n-1}}\leq 2\quad (0)$$

Now we want to show that :

$$(1-x)^{(2x)^n-1}\leq 2(1-x)^{(2x)^{n-1}}\quad(1)$$

For that we need a lemma :

Lemma :

Let $0.5\leq x<1$ and $n\geq 2$ a natural number then we have :

$$f(n)=\ln(1-x)((2x)^n-1-(2x)^{n-1})\leq f(n-1)=\ln(1-x)((2x)^{n-1}-1-(2x)^{n-2})$$

It's true because it's equivalent to :

$$(2x)^{n-2}(2x-1)^2\geq 0$$

So we have :

$$f(n)\leq f(2)$$

We can show also that on $[0.61,1)$ :

$$f(n)\leq f(2)\leq \ln(2)$$

Wich is equivalent to $(1)$

So we have from $(1)$ and $(0)$ we need to show :

$$2x^{(2(1-x))^{n-1}}+2(1-x)^{(2x)^{n-1}}\leq 2$$

Or :

$$x^{(2(1-x))^{n-1}}+(1-x)^{(2x)^{n-1}}\leq 1$$

So now we can use induction to prove $(I)$.

Some idea to go further :

We have the inequality :

$$f(x)=-4\cdot\left(1-x\right)\cdot x\cdot\left(\left(1-x\right)^{\frac{\left(-4\left(1-x\right)x-1\right)}{-2\left(1-x\right)}}\cdot x^{\frac{\left(-4\left(1-x\right)x-1\right)}{-2x}}\right)^{\frac{1}{2}}\left(x^{-\frac{1}{2x}}-\left(1-x\right)^{-\frac{1}{2\left(1-x\right)}}\right)\geq x^{2\left(1-x\right)}-\left(1-x\right)^{2x}\tag{E}$$

On $[0.5,0.6875]$

Now a good idea and a factorization is the expression :

$$\left(f\left(x\right)\right)^{2}+4\left(1-x\right)^{2x}x^{2\left(1-x\right)}$$

let $[0.5,0.55]$:

Then we have :

$$48\left(x-0.5\right)^{2}\geq \left(\left(1-x\right)^{\frac{1}{4\left(1-x\right)}}x^{-\frac{1}{4x}}-\left(1-x\right)^{-\frac{1}{4\left(1-x\right)}}x^{\frac{1}{4x}}\right)^{2}$$

Sketch of proof :

Denotes by :

$$n(x)=\left(1-x\right)^{\frac{1}{4\left(1-x\right)}}x^{-\frac{1}{4x}}$$ $$h\left(x\right)=\left(\left(1-x\right)^{\frac{1}{4\left(1-x\right)}}x^{-\frac{1}{4x}}-\left(1-x\right)^{-\frac{1}{4\left(1-x\right)}}x^{\frac{1}{4x}}\right)$$ And :

$$k(x)=-\left(1-x\right)^{-\frac{1}{4\left(1-x\right)}}x^{\frac{1}{4x}}$$

For $x\in[0.5,0.55]$ we have :

$$\frac{d}{dx}\left(\frac{d}{dx}h\left(x\right)\right)\leq \frac{\frac{d}{dx}\left(\frac{d}{dx}\left(\left(1-x\right)^{\frac{1}{4\left(1-x\right)}}x^{-\frac{1}{4x}}\right)\right)}{\left(1-x\right)^{\frac{1}{4\left(1-x\right)}}x^{-\frac{1}{4x}}}-\frac{\frac{d}{dx}\left(\frac{d}{dx}\left(-\left(1-x\right)^{-\frac{1}{4\left(1-x\right)}}x^{\frac{1}{4x}}\right)\right)}{-\left(1-x\right)^{-\frac{1}{4\left(1-x\right)}}x^{\frac{1}{4x}}}\leq 0$$

So the function $h(x)$ is concave and decreasing .

Remains to use a chord wich lies below the graph of the concave function $h(x)$

$(E)$ is an application of :

Let $-1\leq a<0$ and $0<x\leq y$ then we have :

$$r(x)=a\cdot x^{\frac{\left(a-1\right)}{2}}y^{\frac{\left(a-1\right)}{2}}\left(x-y\right)-x^{a}+y^{a}\geq 0\tag{G}$$

It's not hard to show that $r(x)$ is convex on $(0,y)$ then the tangent of $r(x)$ at $x=y$ lies below the graph of $r(x)$ . The rest is smooth .

Barackouda
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