Nice inequality with exponents $a^{2b}+b^{2a}leq a^{Big(frac{a(1-a)(frac{1}{2}-a)}{4}Big)^2}$

The inequality for $0le ble 1/2$, $a+b=1$, $$a^{2b}+b^{2a}leq a^{Big(frac{ab(b-a)}{8}Big)^2}tag{1}$$ can be strengthened to $$a^{2b}+b^{2a}le e^{-3a^2b^2(b-a)^2/4}tag{2}$$ noting that $e^{-3a^2b^2(b-a)^2/4}le a^{Big(frac{ab(b-a)}{8}Big)^2}$ unless $a$ is almost $0$.

Only a sketch proof is given here, for what it's worth.

For brevity, let $f(x):=x^{1-x}$ and $g(x):=x^2(1-x)^2(x-frac{1}{2})^2$. The claim is $$f(x)^2+f(1-x)^2le e^{-3g(x)}$$

Trivially, $f(x)ge x$, $f(0)=0$, $f'(0)=1$, $f(1)=1$, $f'(1)=0$. $f'(x)=frac{1-x-xln x}{x^x}$.

Proposition 1. $f'(x)=0iff x=1$

Proof: $frac{1}{x}-ln x=1$, equivalent to $frac{1}{x}+lnfrac{1}{x}=1$, so $x=frac{1}{W(e)}=1$. ($W$ is Lambert's function.)

Proposition 2. $f(x)^2+f(1-x)^2$ has three local maxima, at $x=0,frac{1}{2},1$.

Proof: The maxima/minima of $f(x)^2+f(1-x)^2$ occur when $f(x)f'(x)=f(1-x)f'(1-x)$.

At $x=0$, $f(0)=0=f'(1)$; at $x=1$, $f(1-x)=0=f'(1)$. Otherwise, divide by $f(x)$, $f(1-x)$.

At $x=0$, $$f(x)^2+f(1-x)^2=e^{2(1-x)ln x}+e^{2xln(1-x)}=x^2+o(x)+1-2x^2+o(x)=1-x^2+o(x)$$ Hence $x=0$ is a local maximum. By symmetry, so is $x=1$.

The function $frac{f'(x)}{f(1-x)} = frac{1-x-x ln x}{(x(1-x))^x}$ lies between $1$ and $2$, and has one local maximum and one local minimum. A sketch is as follows (blue curve).

$frac{f'(x)}{f(1-x)}=frac{f'(1-x)}{f(x)}$ at three places; the intersections are simple. Since $x=0$ is a local maximum, it follows that the only other local maximum is at $x=1/2$. Then $f(1/2)=1$.

Corollary $F(x):=-ln(f(x)^2+f(1-x)^2)$ also has three local minima at $x=0,frac{1}{2},1$.

A Taylor expansion at each point gives $F(0+h)=h^2+o(h)$, $F(1/2+h)=ch^2+o(h)$ where $c=4 - 4ln2 - 2ln^22approx0.267$, $F(1-h)=h^2+o(h)$.

Hence fitting a polynomial with double roots at $x=0,frac{1}{2},1$, namely $alpha g(x)$, a necessary condition for $alpha g(x)le F(x)$ is $alphale min(4,16c)=4$. A sketch of $F(x)$ shows that the worst cases are at these points; and that $alphale3$ is sufficient for $F(x)ge3g(x)$. No simple proof for this, just a splitting into ranges $[0,1/8]$, $[1/8,3/8]$, $[3/8,1/2]$, and use Taylor series on each.

Proposition 3. For $x>e^{-16alpha}$, $$e^{-alpha g(x)}le x^{(g(x)/4)^2}$$

Proof: Follows from $-alpha g(x)le g(x)ln x/16$, equivalent to $ln xge-16alpha$, or $xge e^{-16alpha}$.