I offer a complete, self-contained solution below that can be checked **without computer/calculator assistance**.

The main tools that we will use (abuse?) are **logarithmic differentiation** and **symmetry**, to greatly simplify the exponentials that we are dealing with.

We may assume that $0<a\leq\frac12\leq b<1$. We split into the following two cases.

# Case 1: $a\leq\frac13$

By Bernoulli's inequality we have $b^{4a^2}=(1-a)^{4a^2}\leq1-4a^3$, so it suffices to show that
$$a^{4(1-a)^2}\stackrel?<4a^3.$$
Let $h(a)=(4(1-a)^2-3)\ln(a)$; we want to show that $h(a)\stackrel?<\ln4$. Now
$$h'(a)=\frac{4(1-a)^2-3}a-8(1-a)\ln(a)$$
can clearly be seen to be decreasing on $a\in[0,1]$ (*edit: in fact this is false as stated, but writing $\frac{4(1-a)^2-3}a=\frac{(2a-1)^2}a-4$ we see that $h'(a)$ is decreasing for $a\in[0,\frac12]$*). Hence
$$h'(a)\geq h'(\frac13)=\frac{16\ln3-11}3>\frac{16-11}3>0,$$
so $h(a)$ is increasing on $a\in[0,\frac13]$. Thus
$$h(a)\leq h(\frac13)=\frac{11}9\ln3<\ln4,$$
since $4^4=256>243=3^5$ implies $\frac{\ln4}{\ln3}>\frac54>\frac{11}9$, as desired.

# Case 2: $\frac13\leq a\leq\frac12$

Substitute $a=\frac12-x$ and $b=\frac12+x$, so $x\in[0,\frac16]$ and
$$b^{4a^2}+a^{4b^2}=\left(\frac12+x\right)^{(1-2x)^2}\!\!\!+\left(\frac12-x\right)^{(1+2x)^2}\!\!\!=F(x)+F(-x),$$
where
$$F(x)=\left(\frac12+x\right)^{(1-2x)^2}\!\!\!.$$

Write $F'(x)=F(x)G(x)$ (so $G$ is the logarithmic derivative of $F$). It is clear that $F(x)$ is increasing on $x\in[-\frac16,\frac16]$, so $F(x),F'(x)>0$ implies $G(x)>0$ on $x\in[-\frac16,\frac16]$.

Now $F(0)+F(-0)=1$, so
\begin{align*}
F(x)+F(-x)\stackrel?\leq1
&\:\Longleftarrow\:\frac d{dx}(F(x)+F(-x))\stackrel?\leq0\\
&\iff F'(x)\stackrel?\leq F'(-x)\\
&\iff\frac{F(x)}{F(-x)}\stackrel?\leq\frac{G(-x)}{G(x)}.
\end{align*}
We now prove this last inequality on $x\in[0,\frac16]$ in the following steps.

## Step 1

We will show that $\ln\left(\dfrac{F(x)}{F(-x)}\right)$ is concave up on $x\in[0,\frac16]$, ie.
\begin{align*}
\frac{d^2}{dx^2}\ln\left(\dfrac{F(x)}{F(-x)}\right)&=\frac d{dx}(G(x)+G(-x))\\
&=G'(x)-G'(-x)\stackrel?\geq0.
\end{align*}
It suffices to show that $G'(x)$ is increasing on $x\in[-\frac16,\frac16]$. Now
\begin{align*}
G(x)&=\frac d{dx}\ln(F(x))\\
&=(1-2x)\left(\frac{1-2x}{\frac12+x}-4\ln\left(\frac12+x\right)\right),\\
G'(x)=\cdots&=\frac43\cdot\frac{-4+9(x+\frac16)^2}{(\frac12+x)^2}+8\ln\left(\frac12+x\right),\tag{*}
\end{align*}
so (by some miracle!) $G'(x)$ can be clearly seen to be increasing on $x\in[-\frac16,\frac16]$.

Hence the graph of $\ln\left(\dfrac{F(x)}{F(-x)}\right)$ lies below the line joining $(0,0)$ and $\left(\dfrac16,\ln\left(\dfrac{F(\frac16)}{F(-\frac16)}\right)\right)$, ie.
$$\ln\left(\dfrac{F(x)}{F(-x)}\right)\leq6\ln\left(\dfrac{F(\frac16)}{F(-\frac16)}\right)x=2Cx,\quad C:=\frac{4\ln54}3.$$

## Step 2

We are left to show that $\dfrac{G(-x)}{G(x)}\stackrel?\geq e^{2Cx}$ on $x\in[0,\frac16]$.

A little manipulation gives
\begin{align*}
\frac{G(x)}{\sqrt{1-4x^2}}&=\sqrt{\frac{1-2x}{1+2x}}\left(2\,\frac{1-2x}{1+2x}-4\ln\left(\frac12+x\right)\right)\\
&=u(2u^2+4\ln(1+u^2)):=H(u),
\end{align*}
under the substitution $u=\sqrt{\dfrac{1-2x}{1+2x}}$, $x=\dfrac12\,\dfrac{1-u^2}{1+u^2}$.

Now the desired inequality is equivalent to
$$\frac{H(\frac1u)e^{-Cx}}{H(u)e^{Cx}}=\frac{G(-x)e^{-Cx}}{G(x)e^{Cx}}\stackrel?\geq1\quad\text{for }x\in[0,\tfrac16]\iff u\in[\tfrac1{\sqrt2},1].$$
Note that $u$ is decreasing in $x$, and the transformation $x\to-x$ is equivalent to $u\to\frac1u$. Thus it is sufficient to show that $H(u)e^{Cx}$ is **increasing** on $u\in[\frac1{\sqrt2},\sqrt2]$ (this is less than clear; I have gotten the sign wrong several times myself), ie.
\begin{align*}
\frac d{du}\ln(H(u)e^{Cx})&=\frac d{du}(Cx+\ln H(u))\\
&=C\frac{dx}{du}+\frac{H'(u)}{H(u)}\\
&=-2C\frac u{(1+u^2)^2}+\frac1u+\frac{2u+\frac{4u}{1+u^2}}{u^2+2\ln(1+u^2)}\stackrel?\geq0\\
\iff\frac{2C}{1+u^2}&\stackrel?\leq\frac{1+u^2}{u^2}+\frac{2(3+u^2)}{u^2+2\ln(1+u^2)}.
\end{align*}
We substitute $t=u^2$ and use the inequality $\ln(1+t)\leq\ln2+\frac{t-1}2$ (by convexity of $\ln(t+1)$; RHS is tangent line at $t=1$) to reduce the above to
$$2+\frac1t+\frac{7-2\ln2}{2t+2\ln2-1}\stackrel?\geq\frac{2C}{1+t}.\tag{**}$$
This will fall to Cauchy-Schwarz (in the Engel form $\sum\dfrac{a_i^2}{b_i}\geq\dfrac{(\sum a_i)^2}{\sum b_i}$), if we can find the correct weights. With some (okay, a lot) of inspiration, we obtain the following:
\begin{align*}
&\phantom{{}={}}2+\frac1t+\frac{7-2\ln2}{2t+2\ln2-1}\\
&\geq\frac{94}{47}+\frac{18}{18t}+\frac{101}{36+7t}\\
&\geq\frac{(\sqrt{94}+\sqrt{18}+\sqrt{101})^2}{54(1+t)}\geq\frac{2C}{1+t},
\end{align*}
which is what we wanted.

Hence we are left with the following (zero-variable!) inequalities that we used above:
$$\ln2\stackrel?\leq\frac{25}{36},\qquad\frac{(\sqrt{94}+\sqrt{18}+\sqrt{101})^2}{54}\stackrel?\geq2C.$$
Most people should be content with checking these by calculator. For the purists, here is a sketch of how to get these by hand.

## Step 3 (Optional?)

Firstly, note that for $x>0$, we have
\begin{align*}
\ln\left(\frac{1+x}{1-x}\right)&=2\left(x+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\cdots\right)\\
&\leq2\left(x+\frac{x^3}3+\frac{x^5}3+\frac{x^7}3+\cdots\right)\\
&=2\left(x+\frac{x^3}{3(1-x^2)}\right).
\end{align*}
Taking $x=\frac13$ gives the first inequality. In addition, taking $x=\frac15$ gives
$$54(2C)=54\times\frac83\left(3\ln\frac32+\ln2\right)\leq72\left(3\times\frac{73}{180}+\frac{25}{36}\right)=\frac{2876}5.$$
Let $(a,b,c,d)=(101,94,18,\frac{2876}5)$, and $n=\frac12(c+d-a-b)$. We are left to check that
\begin{align*}
\sqrt a+\sqrt b+\sqrt c&\geq\sqrt d\\
\iff a+b+2\sqrt{ab}&\geq c+d-2\sqrt{cd}\qquad(\because\sqrt a+\sqrt b>0)\\
\iff\sqrt{ab}&\geq n-\sqrt{cd}\\
\iff ab&\geq n^2+cd-2n\sqrt{cd}\qquad(\because\sqrt{ab}>0)\\
\iff2n\sqrt{cd}&\geq n^2+cd-ab\\
\iff4n^2cd&\geq(n^2+cd-ab)^2,\qquad(\because2n\sqrt{cd}>0)
\end{align*}
which after substitution becomes $\dfrac{205212545208}{125}\geq\dfrac{16402832101681}{10000}$, something that I consider to be close to the limits of what I can do by hand.

**QED**. Phew!

## Remarks

Yes, this might be a long proof, and perhaps it doesn't give much insight as to *why* the original inequality holds. However, I hope that the tricks used in the solution (and there are many!) are of independent interest.

I have optimised the solution as presented above to minimise the effort of hand calculations. Some parts of the solution have feasible alternatives, eg. at $(*)$ we can compute $G''(x)=\dfrac{16(7+8x+4x^2)}{(1+2x)^3}$, and check that the numerator is positive on $x\in[-\frac16,\frac16]$; at $(**)$ we essentially need $P(t)\geq0$ for some cubic $P$, and we can proceed by showing
$$P(t)\geq k(t-\lambda)(t-\mu)^2+Q(t)$$
for good choices of $k,\lambda,\mu\in\mathbb R$ and $Q$ a quadratic in $t$.

**How human is this solution?** It is certainly human-*checkable*, but I have serious doubts as to whether a human can *come up* with a proof along these lines without computer assistance. As many previous answers have noted, the original inequality is very tight, and we cannot afford to lose more than $0.03$ in total throughout our proof.

In addition, I can count about 8 places in Step 2 alone where I used the irreversible $\Leftarrow$ implication, ie. "to prove A it is sufficient to prove B." This is disastrous if B turns out to be false! I personally needed extensive computer aid to explore the problem space and avoid dead ends of this type --- this proof was made possible by around 2000--3000 Mathematica commands.

For these two reasons, I doubt that there can be an unassisted proof using the normal tools of calculus (such as those in the above solution), let alone in a competition setting. Of course, maybe we just need some radical new insight or perspective. (I have not looked at the $W$ function closely, so maybe...?)

Cheers.