If $0I 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 :)
 A: 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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A: We want to show 1:

Let $0<x<0.5$ 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 $0<x\leq 0.5$:
$$x^{2\left(1-x\right)}\leq \frac{\left(\left(\frac{1}{2x}-1\right)\left(\frac{2}{\left(4x^{2}+1\right)}\right)+1\right)^{-1}}{2^{\left(2-2x\right)}}$$
