How to prove this integral inequality $ \int_0^{2\pi} p(x)[p(x)+p''(x)] dx \int_0^{2\pi}\frac{1}{p(x)+p''(x)} dx\geq 2\pi \int _0^{2\pi} p(x) dx $?

Question

Let $p\in C^2(\mathbb{R})$ be a $2\pi$-periodic function such that $p(x)>0$ and $p(x)+p''(x)>0$ for all $x\in \mathbb{R}$. Then it holds
$$ \int_0^{2\pi} p(x)[p(x)+p''(x)] dx \int_0^{2\pi}\frac{1}{p(x)+p''(x)} dx\geq 2\pi \int _0^{2\pi} p(x) dx $$ The identity holds when $p+p''$ is a constant.

Answer 1

I can verify this for $$p(x)=a_0 \cos(0x)+a_1 \cos(1x) + a_2\cos(2x).$$

By hypothesis, $p(x)+p''(x)$ is positive. So that sum is positive for $x=0$ and $x=\pi/2$, which means that $a_0+3a_2$ and $a_0-3a_2$ are both positive. Averaging these, $a_0$ is positive. Multiplying these, $a_0^2-9a_2^2$ is positive. Dividing by three and adding a square, $2a_0^2-3a_2^2$ is also positive.

Now when we evaluate all the integrals, the inequality becomes $$\big[\pi(2a_0^2-3a_2^2)\big]\big[\frac{2\pi}{\sqrt{a_0^2-9a_2^2}}\big]\ge 2\pi \big[2\pi a_0\big].$$

Equivalently $$\frac{2a_0^2-3a_2^2}{\sqrt{a_0^2-9a_2^2}}\ge 2a_0.$$

Since the numerator, denominator, and right-hand side are all positive, this is equivalent to $$\frac{(2a_0^2-3a_2^2)^2}{a_0^2-9a_2^2}\ge 4a_0^2.$$ That in turn simplifies to $$24a_0^2 a_2^2+9a_2^4 \ge 0,$$ which is true. Can we generalize this to other periodic $p$, probably by using Holder's inequality?