Monstruous inequality $\Big(a^{(4b^2)^{4a^2}}+b^{(4a^2)^{4b^2}}\Big)^{\frac{1}{2}}\Big(a^{(4b^2)^{2b}}+b^{(4a^2)^{2a}}\Big)^{\frac{1}{2}}\geq 1$

Question

Hi I want to solve this ,

Let $a,b>0$ such that $a+b=1$ then we have : $$\Big(a^{(4b^2)^{4a^2}}+b^{(4a^2)^{4b^2}}\Big)^{\frac{1}{2}}\Big(a^{(4b^2)^{2b}}+b^{(4a^2)^{2a}}\Big)^{\frac{1}{2}}\geq 1$$

I have try to replace $a$ and $b$ by $1-x$ and $x$ to see how monstruous it is ...

With wolfram alpha i try to calculate the power series but it becomes too complexe for me .

I try also trigonometry but it's awful .

With lucidity we need a book to solve this .

The equality occurs for $a=b=0.5$

Thanks a lot for sharing your time and knowledge