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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. enter image description here

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    For what it's worth, you may assume $b=1$ due to scale invariance of the inequality.2011-06-06

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