If $a+b=1$ then $a^{4b^2}+b^{4a^2}\leq1$ Let $a$ and $b$ be positive numbers such that $a+b=1$. Prove that:
$$a^{4b^2}+b^{4a^2}\leq1$$
I think this inequality is very interesting because the equality "occurs" for $a=b=\frac{1}{2}$ and also for $a\rightarrow0$ and $b\rightarrow1$.
I tried to work with a function of one variable, but the derivative is not easy.
I also don't get something solvable by Taylor series.
 A: Update 
I want to share some thought. Consider the more general problem
$$
a^{n b^2} + b^{n a^2} \leq 1
$$
A key observation is the symmetry of the two terms $a^{n b^2} $ and $b^{n \
a^2} $. Due to the constraint $a + b = 1 $, 
the two terms are just $a^{n (1 - a)^2} $ and $ (1 - 
      a)^{n a^2} $. So a substitution of $a \rightarrow 1 - 
  a$ changes one term to the other. 
    Conclusion : The LHS is a sum of two terms symmetric around the $a = 
 1/2 $.
Lemma W.l.o.g, suppose a function $f (x) $ in interval $[0, 
  1] $.If the function is monotonic and convex around $x=1/2$, 
then the "mirror average function" $g (x) = (f (x) + f (1 - x))/2 $ has a maximum at $x = 
  1/2 $.
Proof Just calculate to show $g' (1/2) = 0 $ and $g'' (x) = f'' (x)$
Corollary for concave $f(x)$ follows directly. This analysis does not answer the question, but hopefully will introduce some abstractness and shed more light on it.
Old Post
This should be a comment, but then I won't be able to post pictures. For the more general inequality:
$$
a^{n b^2}+b^{n a^2}\leq1
$$
I draw pictures for n = 0, 1, ..., 7. Each one has green dashed contour highlighting where the equality is satisfied. And of course, each one is overlaid with $a+b=1$. It's interesting to note only $n = 4$ is tightly bounded by the green contour, so it is a really special $n$ value.

A: This is indeed a hard nut, since convexity cannot be invoked to close the case. The following plot shows that the function
$$f(x):=(1-x)^{4x^2}+x^{4(1-x)^2}\qquad(0\leq x\leq 1)$$
in fact never drops below $0.97\>$! (Compare Robin Aldabanx' answer)

At the moment I'm just able to show that $f(x)$ behaves as claimed near $x=0$ (and, by symmetry, near $x=1$) and near $x={1\over2}$.
If $0\leq x\leq{1\over2}$ then $0\leq4x^2\leq1$, and Bernoulli's inequality gives
$$(1-x)^{4x^2}\leq1-4x^3\ .$$
On the other hand
$$x^{4(1-x)^2}=x^4\cdot x^{-8x+4x^2}=:x^4\> h(x)$$
with $\lim_{x\to0+}h(x)=1$. It follows that there is a $\delta>0$ with
$$f(x)\leq 1-4x^3+2x^4=1-4x^3\>\left(1-{x\over2}\right)<1\qquad (0<x<\delta)\ .$$
For $x\doteq{1\over2}$ we consider the auxiliary function
$$g(t):=f\left({1\over2}+t\right)\qquad\bigl(|t|\ll1\bigr)$$
which is analytic for small $|t|$. Mathematica computes its Taylor series as
$$g(t)=1+\left(-8+4\log 2+8\log^2 2\right) t^2+\ ?\>t^3\ .$$
The numerical value of the relevant coefficient here is $\doteq-1.38379$, and this tells us that $f$ has  a local maximum at $x={1\over2}$.
A: Again too long for  comment
The inequality is equivalent to on $(0,0.5)$:
$$f(x)=\left(\left(x^{-\left(2\left(1-x\right)\right)^{2}}\right)-1\right)\left(\left(1-x\right)^{-\left(2x\right)^{2}}-1\right)\geq1$$
Now define :
$$u(x)=1+\frac{\left(\sqrt{1+0.0025\left(\frac{\left(\ln\left(m\left(x\right)\right)\right)}{\ln\left(2\right)}\right)^{\frac{\ln\left(64\right)}{3}}\cdot\left(\ln\left(\frac{m\left(x\right)}{2}\right)\right)^{2}}\right)-1}{0.00125\left(\ln\left(\frac{m\left(x\right)}{2}\right)\right)^{2}}$$
Where $m(x)=x^{-\left(2\left(1-x\right)\right)^{2}}$ and $g\left(x\right)=\left(\left(1-x\right)^{-\left(2x\right)^{2}}-1\right)$
We have the inequalities :
$$f(x)+(u(x)-1)g(x)\geq 2$$
$$f(x)\geq \frac{\left(\left(u\left(1-x\right)\right)\cdot\left(u\left(x\right)\right)\right)^{0.25}}{\sqrt{2}}\geq 1$$
This kind of method is valid for a generalisation when the higher exponent (here $2$ in $f(x)$) is greater or equal to $2$.
