The isoperimetric theorem states that the area A of a plane region with perimeter P cannot exceed $\pi ({P/2\pi})^2\ = P^2/4\pi.$
That is, $P \geqslant \sqrt{4\pi A} $
Considering an ellipse with major and minor semi-axes a and b respectively, its area is $\pi ab$
and its perimeter is given by $\int\limits_{0}^{2\pi}\sqrt{a^2sin^2t + b^2cos^2t}\ dt$
So we have $\int\limits_{0}^{2\pi}\sqrt{a^2sin^2t + b^2cos^2t}\ dt \geqslant \sqrt{4\pi(\pi ab)}$
But by quartering and rearranging the elllpse as in the diagram below, with a central square of area $(a - b)^2 $, we can state the stronger
$\int\limits_{0}^{2\pi}\sqrt{a^2sin^2t + b^2cos^2t}\, dt \geqslant \sqrt{4\pi[\pi ab + (a - b)^2]}$
And my question is whether this last inequality can be proven analytically.