inequalities - Is there a good reason why a^{2b} + b^{2a}

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:
$$
begin{aligned}
a^{2b}&=(1-b)^{1-t}=1-b(1-t)-t(1-t)left[frac{1}2b^2+frac{1+t}{3!}b^3+frac{(1+t)(2+t)}{4!}b^4+dotsright]
\
&le 1-b(1-t)-t(1-t)left[frac{b^2}{1cdot 2}+frac{b^3}{2cdot 3}+frac{b^4}{3cdot 4}+dotsright]
\&
=1-b(1-t)-t(1-t)left[blogfrac 1{a}+b-logfrac {1}aright]
\
&=1-b(1-t^2)+(1-b)t(1-t)logfrac{1}a=1-bleft(1-t^2-t(1+t)logfrac 1aright)
end{aligned}
$$
(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 $uge 0$.
For $k=1$ it is just $e^uge 1+u+frac{u^2}{2}$. For $kge 2$, 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 $frac{k^0}{0!}=1$, $frac{k^j}{j!}ge frac {k^j}{j^{j-1}}ge kge 2$ for $1le jle k$, and $frac{k^{k+1}}{(k+1)!}ge frac{k}{k+1}$.

Step 3:
Let $u=logfrac 1a$. We've seen in Step 1 that $a^{2b}le 1-b(1-tmu)$ where $mu=u+(1+u)t$. In what follows, it'll be important that $mulefrac 1a-1+frac 1a t=1$ (we just used $logfrac 1ale frac 1a-1$ here.

We have $b^{2a}=b(a-t)^t$. Thus, to finish, it'll suffice to show that $(a-t)^tle 1-tmu$. Taking negative logarithm of both sides and recalling that $frac 1a=e^u$, we get the inequality
$$
tu+tlog(1-te^u)^{-1}ge log(1-tmu)^{-1}
$$
to prove.
Now, note that, according to Step 2,
$$
begin{aligned}
&frac{e^{uk}}kge frac{(1+u)(1+u+dots+u^{k-1})}k+frac{u^{k+1}}{k+1}
gefrac{(1+u)(mu^{k-1}+mu^{k-2}u+dots+u^{k-1})}k+frac{u^{k+1}}{k+1}
\
&=frac{mu^k-u^k}{kt}+frac{u^{k+1}}{k+1}
end{aligned}
$$
Multiplying by $t^{k+1}$ and adding up, we get
$$
tlog(1-te^u)^{-1}ge -ut+log(1-tmu)^{-1}
$$
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 $xmapsto (1+x)^b-(1-x)^a$ 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 $(1+x)^b-(1-x)^age x$ on $[0,1]$. Plugging in $x=t$ finishes the story.