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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. –  Jonas Meyer Jun 6 '11 at 2:20
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1 Answer 1

We have to prove that $$\int_0^{\pi/2}\sqrt{a^2\sin^2 t+b^2\cos^2 t}dt\geq{\pi\over2}\sqrt{ab+(a-b)^2/\pi}\ .$$ The following sketch of a proof is not beautiful because convexity isn't helping us. We may assume $a^2+b^2=2$ and write $$a=\sqrt{2}\cos({\pi\over4}-\alpha)\ ,\quad b=\sqrt{2}\cos({\pi\over4}+\alpha)\qquad(0\leq\alpha\leq{\pi\over4}).$$ Then $$(a-b)^2=4\sin^2\alpha,\qquad ab =1-2\sin^2\alpha,$$ and the right hand side of the stated inequality becomes $$RHS={\pi\over2}\sqrt{1-(2-4/\pi)\sin^2\alpha}\leq {\pi\over2}-({\pi\over2}-1)\sin^2\alpha.\qquad(1)$$ On the other hand, $$a^2\sin^2 t+b^2\cos^2 t=1-\sin(2\alpha)\cos(2t)$$ and so the left hand side of the stated inequality is seen to be equal to $$LHS=\int_0^{\pi/2}{1\over2}\Bigl(\sqrt{1-\sin(2\alpha)\cos\tau}+\sqrt{1+\sin(2\alpha)\cos\tau}\Bigr)\ d\tau.$$ Now by looking at a Mathematica plot one gets the impression that $$f(x):={1\over2}\bigl(\sqrt{1-x}+\sqrt{1+x}\bigr)\geq 1-{2x^2\over 15}-{x^4\over6}=:g(x)\qquad(0\leq x\leq 1),$$ from which one derives $$LHS\geq \int_0^{\pi/2} g\bigl(\sin(2\alpha)\cos\tau\bigr)\ d\tau={\pi\over2}-{\pi\cos^2\alpha\over240}(47-15\cos\alpha)\sin^2\alpha.\qquad(2)$$ Mathematica again shows that the coefficients of $\sin^2\alpha$ in (1) and (2) are related by $${\pi\cos^2\alpha\over240}(47-15\cos\alpha)\leq {\pi\over2}-1\qquad(0\leq\alpha\leq{\pi\over4}).$$

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