Is there a good reason why $a^{2b} + b^{2a} \le 1$ when $a+b=1$? The following problem is not from me, yet I find it a big challenge to give a nice (in contrast to 'heavy computation') proof. The motivation for me to post it lies in its concise content.
If $a$ and $b$ are nonnegative real numbers such that $a+b=1$, show that  $a^{2b} + b^{2a}\le 1$.
 A: Fixed now. I spent some time looking for some clever trick but the most unimaginative way turned out to be the best. So, as I said before, the straightforward Taylor series expansion does it in no time. 
Assume that $a>b$. Put $t=a-b=1-2b$.
Step 1: 
$$
\begin{aligned}
a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)\left[\frac{1}2b^2+\frac{1+t}{3!}b^3+\frac{(1+t)(2+t)}{4!}b^4+\dots\right]
\\
&\le 1-b(1-t)-t(1-t)\left[\frac{b^2}{1\cdot 2}+\frac{b^3}{2\cdot 3}+\frac{b^4}{3\cdot 4}+\dots\right]
\\&
=1-b(1-t)-t(1-t)\left[b\log\frac 1{a}+b-\log\frac {1}a\right]
\\
&=1-b(1-t^2)+(1-b)t(1-t)\log\frac{1}a=1-b\left(1-t^2-t(1+t)\log\frac 1a\right)
\end{aligned}
$$
(in the last line we rewrote $(1-b)(1-t)=(1-b)2b=b(2-2b)=b(1+t)$)
Step 2.
We need the inequality $e^{ku}\ge (1+u)(1+u+\dots+u^{k-1})+\frac k{k+1}u^{k+1}$ for $u\ge 0$.
For $k=1$ it is just $e^u\ge 1+u+\frac{u^2}{2}$. For $k\ge 2$, the Taylor coefficients on the left are $\frac{k^j}{j!}$ and on the right $1,2,2,\dots,2,1$ (up to the order $k$) and then $\frac{k}{k+1}$. Now it remains to note that $\frac{k^0}{0!}=1$, $\frac{k^j}{j!}\ge \frac {k^j}{j^{j-1}}\ge k\ge 2$ for $1\le j\le k$, and $\frac{k^{k+1}}{(k+1)!}\ge \frac{k}{k+1}$.
Step 3:
Let $u=\log\frac 1a$. We've seen in Step 1 that $a^{2b}\le 1-b(1-t\mu)$ where $\mu=u+(1+u)t$. In what follows, it'll be important that $\mu\le\frac 1a-1+\frac 1a t=1$ (we just used $\log\frac 1a\le \frac 1a-1$ here.
We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^t\le 1-t\mu$. Taking negative logarithm of both sides and recalling that $\frac 1a=e^u$, we get the inequality
$$
tu+t\log(1-te^u)^{-1}\ge \log(1-t\mu)^{-1}
$$
to prove.
Now, note that, according to Step 2, 
$$
\begin{aligned}
&\frac{e^{uk}}k\ge \frac{(1+u)(1+u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1}
\ge\frac{(1+u)(\mu^{k-1}+\mu^{k-2}u+\dots+u^{k-1})}k+\frac{u^{k+1}}{k+1}
\\
&=\frac{\mu^k-u^k}{kt}+\frac{u^{k+1}}{k+1}
\end{aligned}
$$
Multiplying by $t^{k+1}$ and adding up, we get
$$
t\log(1-te^u)^{-1}\ge -ut+\log(1-t\mu)^{-1}
$$
which is exactly what we need.
The end.
P.S. If somebody is still interested, the bottom line is almost trivial once the top line is known. Assume again that $a>b$, $a+b=1$. Put $t=a-b$.
$$
\begin{aligned}
&\left(\frac{a^b}{2^b}+\frac{b^a}{2^a}\right)^2=(a^{2b}+b^{2a})(2^{-2b}+2^{-2a})-\left(\frac{a^b}{2^a}-\frac{b^a}{2^b}\right)^2
\\
&\le 1+\frac 14\{ [\sqrt 2(2^{t/2}-2^{-t/2})]^2-[(1+t)^b-(1-t)^a]^2\}
\end{aligned}
$$
Now it remains to note that $2^{t/2}-2^{-t/2}$ is convex on $[0,1]$, so, interpolating between the endpoints, we get $\sqrt 2(2^{t/2}-2^{-t/2})\le t$. Also, the function $x\mapsto (1+x)^b-(1-x)^a$ is convex on $[0,1]$ (the second derivative is $ab[(1-x)^{b-2}-(1+x)^{a-2}]$, which is clearly non-negative). But the derivative at $0$ is $a+b=1$, so $(1+x)^b-(1-x)^a\ge x$ on $[0,1]$. Plugging in $x=t$ finishes the story. 
A: Since this question has been bumped up I would like to state what I think is its natural framework:  We have the inequality  $f(a,b) \leq 1$
 for $a+b=k$ when $k$ lies between $\frac 12$ and $1$.  Otherwise, the inequality is $f(a,b) \leq \frac {k^k}{2^{k-1}}$ (with $f(a,b) = a^{2b}+b^{2a}$).  This version is not just more comprehensive but it illustrates the dichotomy in where the maximum occurs (at the symmetric point $(\frac k2,\frac k2)$ or at the boundary $(k,0)$).  The two cases considered above ($k=\frac 12$ and $k=1$) are precisely the transitional ones. One can also get estimates from below   (usually by the constant $1$  but in a small neighbourhood around the critical interval $[\frac 12,1]$ the sharp version involves  values which are given implicitly as the solution of transcendental equations).
(P.S.  I had already given some of this information in a comment but, since it elicited no reaction, I have taken the liberty to repeat it here despite the fact that it isn't really an answer but, hopefully, does shed some light on the problem and its solution).
A: By symmetry, it is enough to show that 
\begin{equation}
 f(x):=x^{2-2 x}+(1-x)^{2 x}\le1
\end{equation}
for 
\begin{equation}
 x\in(0,\tfrac12);  
\end{equation}
the latter condition will be assumed by default. 
It is easy to see that 
\begin{equation}
 r_0(x):=1 - x - x \ln x>0. 
\end{equation}
So, $f'(x)$ equals 
\begin{equation}
 f_1(x):=\frac{f'(x)}{x^{1 - 2 x} r_0(x)}
\end{equation}
in sign. Moreover, 
\begin{equation}
 f_2(x):=f'_1(x)\frac{r_0(x)^2}{2 \big((1 - x) x\big)^{2 x}} 
\end{equation}
is a rational function of $x,\ln x,\ln(1-x)$, and $f_2(x)$ equals $f'_1(x)$ in sign. 
This almost completes the solution, since the sign pattern of such a rational function can be determined completely algorithmically, and then one can go back and determine the sign patterns of $f_1$ and $f-1$, in this order. So, the only remaining problem is to find a better algorithm. 
Differentiating $f_2(x)$, we kill the terms containing $\ln^2 x\ln(1-x)$ and $\ln x\ln^2(1-x)$, so that 
\begin{equation}
 f_3(x):=f'_2(x)/c_3(x)
\end{equation}
is a polynomial of degree 1 in $\ln x,\ln^2 x,\ln(1-x),\ln^2(1-x),\ln x\ln(1-x)$ (with coefficients in $\mathbb Z(x)$), where 
\begin{equation}
 c_3(x):=\frac{4 \left(2 x^3-3 x^2-x+1\right)}{(1-x)^2 x^2}>0. 
\end{equation}
Next, 
\begin{equation}
 f_4(x):=f'_3(x)\,\frac{(1 - x)^3 x^3 c_3(x)^2}{4 - x + x^2}
\end{equation}
is a polynomial of degree 1 in $\ln x,\ln^2 x,\ln(1-x),\ln^2(1-x)$ 
and 
\begin{equation}
 f_5(x):=f'_4(x)\,\frac{x^2 (4 - x + x^2)^2}{2c_3(x)(1-x)c_5(x)} 
\end{equation}
is a polynomial of degree 1 in $\ln x,\ln(1-x)$, where 
$c_5(x):=12 + 4 x + 47 x^3 - 99 x^4 + 10 x^5 - 6 x^6>0$.  
Further, 
\begin{equation}
 f_6(x):=f'_5(x)\, \frac{c_5(x)^2 (1 - x)^4}{x (4 - x + x^2)}
\end{equation}
is a polynomial of degree 1 in $\ln x$, and hence so is 
$f'_6(x)$. Therefore, the value of $f'_6(x)$ is between those of $f_{73}(x)$ and $f_{74}(x)$, where $f_{7k}(x)$ is obtained from $f'_6(x)$ by replacing there $\ln x$ with the Taylor polynomial of degree $k$ for $\ln x$ at the point $1/2$. 
Since $f_{73}(x)$ and $f_{74}(x)$ are rational expressions (in $\mathbb R(x)$), it is easy to see that $f_{73}<0$, $f_{74}<0$, and hence $f'_6<0$, so that $f_6$ decreases (on $(0,1/2)$). 
Moreover, $f_6(0+)>0>f_6(1/2)$. So, $f_6$ is $+-$; that is, it changes in sign only once on $(0,1/2)$, from $+$ to $-$. 
So, $f_5$ is up-down; that is, it increases on $(0,c)$ and decreases on $(c,1/2)$, for some $c\in(0,1/2)$. 
Also, $f_5(0+)=0=f_5(1/2)$. So, $f_5>0$, $f_4$ increases, to $f_4(1/2)=0$. 
So, $f_4<0$, $f_3$ decreases, from $f_3(0+)=0$. 
So, $f_3<0$, $f_2$ decreases, from $f_2(0+)=\infty$ to $f_2(1/2)=-1<0$. 
So, $f_2$ is $+-$, $f_1$ is up-down, with $f_1(0+)=-2<0$ and $f_1(1/2)=0$. 
So, $f_1$ is $-+$, $f$ is down-up, with $f(0+)=f(1/2)=1$. 
So, $f<1$ on $(0,1/2)$. QED
A: I think it gives a better sense of the geometry of the problem to ask whether, with non-negative $x,y$ such that $$ \frac{1}{2} \leq x + y \leq 1, $$ we can prove that $$ x^{2 y} + y^{2 x} \leq 1 ?$$ I'm not entirely certain where the second level curve  component, through $\left( \frac{1}{4} , \frac{1}{4}\right),$ meets the axes. My programmable calculator seems to think that, if this arc does have $\left( \frac{1}{2} , 0 \right)$ as a limit point, the arc is tangent to the $x$-axis. 
I see, this was pointed out in a comment on March 17 by Yaakov Baruch, one needs to click on the "show 6 more comments." I think I will leave this here anyway.
A: This is too long to be a comment.
This inequality appears as conjecture 4.8 in this article here. As you probably know, V.Cirtoaje has written many books on olympiad-style inequalities, so you see my reason for not believing that a simple solution exists. Optimization problems can sometimes (or most of the time actually) require "non-elegant" analysis (whatever that means to you) so this search is a bit pointless in my opinion. If an elegant solution is found to some nontrivial optimization/estimation problem then it is very likely to appear in an olympiad/competition, and AOPS is the right place to carry such discussions.
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
A: Maybe there is a small trick yet.
For $a + b = 1$ we can write the sum as 
$$a^{2(1-a)} + b^{2(1-b)} = (\frac{a}{a^a})^2+ (\frac{b}{b^b})^2 $$.
Obviously the sum is 1 for $(a, b) = (0,1), (\frac{1}{2},\frac{1}{2}), (1, 0)$. The question is whether at $(\frac{1}{2},\frac{1}{2})$ there is a maximum or a minimum, i.e. whether the second derivative of
$$(\frac{x}{x^x})^2+ (\frac{1-x}{(1-x)^{1-x}})^2$$
is positive or negative. (By symmetry only minimum or maximum can occur there.)
Using the fact that, for $x = \frac{1}{2}$,
$$\frac{d^2}{dx^2}(\frac{1-x}{(1-x)^{1-x}}) =  \frac{d^2}{(-dx)^2}(\frac{x}{x^x})^2 =  \frac{d^2}{dx^2}(\frac{x}{x^x})^2$$
it is sufficient to prove
$$2\frac{d^2}{dx^2}(\frac{x}{x^x})^2 < 0$$
for $x = \frac{1}{2}$ which in fact is easily demonstrated.
A: This type of problem can be solved by the following approach:


*

*Maximize the function a^(2b)+b^(2a) s.t. a+b=1.  

*We find that the function is maximized at a=b=1/2 and takes value 1.

