# If $0<a<1, 0<b<1$, $a+b=1$, then prove that $a^{2b}+ b^{2a} \le 1$

I have been really struggling with this problem ... please help! Let a,b be real numbers. If $0<a<1, 0<b<1, a+b=1$, then prove that $a^{2b} + b^{2a} \le 1$

What I have thought so far: without loss of generality we can assume that $a \le b$, since $a^{2b} + b^{2a}$ is symmetric in $a$ and $b$. This gives us $0<a \le 1/2, 1/2 \le b<1$. But then I am stuck. I also thought of solving by Lagrange's multiplier method, but it produces huge calculations.

Any help is welcome :)

• This very question cropped up on mathoverflow a few weeks ago. May 18, 2013 at 20:53
• @Olivier: Can you give the link to the mathoverflow question page? May 18, 2013 at 20:57
• mathoverflow.net/questions/17189/… from 2010. May 18, 2013 at 21:01
• Looking at the MO thread...what kind of professor left such problems as homework? Maybe it is because you arrived at the classroom late like George Dantzig did (en.wikipedia.org/wiki/George_Dantzig) and your professor happened to have the habit of leaving hard problems on the blackboard like Prof Neyman? May 18, 2013 at 22:02

Not having a good day with websites. I have downloaded what seems to be the source of the question, a 2009 paper by Vasile Cirtoaje which is about 14 pages. Then a short answer, in a four page document by Yin Li, probably from the same time or not much later. The question was posted on MO by a selfish guy who knew the status of the problem but was hoping for a better answer, a complete answer was also given there in 2010 by fedja, https://mathoverflow.net/questions/17189/is-there-a-good-reason-why-a2b-b2a-1-when-ab1

I have both pdfs by Cirtoaje and Li, email me if you cannot find them yourself.

This is not a reasonable homework question, so I would like to know more of the story, what course for example.

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Yin Li of Binzhou University, Shandong, 2009 or 2010, excerpts I believe he just spelled Jensen's incorrectly, see JENSEN

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• This is not actually a homework problem, my teacher found this problem from some other student and tried for himself but could not solve it, so he gave it to me to have a look. Neither he, nor I had any idea about its level, since it is so simple looking. Now I have clear idea about its hardness, though :). May 19, 2013 at 4:31

We want to show 1:

Let $$0 such that then we have : $$f(x)=x^{2(1-x)}+(1-x)^{2x}\leq q(x)=(1-x)^{2x}+2^{2x+1}(1-x)x^2\leq 1$$

The Lhs is equivalent to :

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

Or : $$\ln\Big(x^{2(1-x)}\Big)\leq \ln\Big(2^{2x+1}(1-x)x^2\Big)$$

Making the difference of these logarithm and introducing the function :

$$g(x)=\ln\Big(x^{2(1-x)}\Big)-\ln\Big(2^{2x+1}(1-x)x^2\Big)$$

The derivative is not hard to manipulate and we see that it's positive and $$x=0.5$$ is an extrema .The conclusion is :

$$g(x)\leq g(0.5)=0$$

And we are done with the LHS.

For the Rhs I use one of the lemma (7.1) due to Vasile Cirtoaje we have :

$$(1-x)^{2x}\leq p(x)=1-4(1-x)x^{2}-2(1-x)x(1-2 x)\ln(1-x)$$

So we have :

$$q(x)\leq p(x)+h(x)$$

We want to show that :

$$p(x)+h(x)\leq 1$$

Wich is equivalent to :

$$-2(x-1)x((4^x-2)x+(2x-1)\ln(1-x))\leq 0$$

It's not hard so I omitt here the proof of this fact .

We are done .

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

Bonus inequality for $$0:

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