# Isoperimetric inequality transformed to an integral inequality

In the following article, it is shown that Gage's inequality (a form of isoperimetric inequality) can be generalised as the following integral inequality which is presented as a conjecture in the article:

Suppose $p(\theta)$ is a $C^2$ periodic function with period $2\pi$ with $p(\theta)>0$ and $p(\theta)+p''(\theta)>0$. Then

$$\int_0^{2\pi}p(\theta)(p(\theta)+p''(\theta))d\theta \int_0^{2\pi}\frac{d\theta}{p(\theta)+p''(\theta)}\geq 2\pi \int_0^{2\pi}p(\theta)d\theta$$

Since the inequality is presented there as a conjecture, I guess it cannot be attacked by regular integral inequalities. I tried to solve it but didn't have any progress with it.

 The question is to prove the inequality.

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What is the question? –  Aryabhata Jun 2 '11 at 8:39
Is is simply asking to show that the LHS is $\geq$ to the RHS? –  night owl Jun 2 '11 at 8:50