Prove that $0\le\int_0^1\log(u){\rm d}x+\frac1{2\pi^2}\int_0^1\frac1{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x$ I want to prove a special case of a functional inequality stated in a book. To be specific, let $u(x)$ being a positive, differentiable function on $[0,1]$ with unit mass (i.e., $\int_0^1 u(x){\rm d}x = 1$), then we want to show that
$$0 \leq \int_0^1 \log(u){\rm d}x + \frac{1}{2\pi^2}\int_0^1 \frac{1}{u^2}\left(\frac{{\rm d}u}{{\rm d}x}\right)^2{\rm d}x.$$
In a book, the author resorts to the construction of certain solutions of one-dimensional heat equation and used the eigen-expansion (hence the arguments are very lengthy and involved). Since I am only interested in this special inequality stated above, may I know whether there exists a simpler and more elegant proof ?

OK, after looking back at the various different proofs of the classical log Sobolev inequality, there is at least one way we can vanquish this. It turns out it is just a special case of the inequality stated (and proved) in section 2.1. of the lecture notes "Entropy Methods and Related Functional Inequalities" written by Daniel Matthes. So here is the hammer: 
Assume that $\phi:\mathbb{R}_+ \to \mathbb{R}$ is convex such that $(\phi'')^{-\frac{1}{2}}$ is concave, let $\psi$ be such that $(\psi'(s))^2 = \phi''(s)$. Then we have that
$$ \int_0^1 \phi(u){\rm d}x - \phi\left(\int_0^1 u{\rm d}x\right) \leq \frac{1}{2\pi^2}\int_0^1 (\psi(u)_x)^2{\rm d}x \quad (*)$$
holding for all smooth, positive functions $u$ on $[0,1]$. Notice that if $\phi(s) = \frac 12 s^2$ and $\psi(s) = s$, $(*)$ becomes the usual Poincare inequality
$$\int_0^1 u^2{\rm d}x - \left(\int_0^1 u{\rm d}x\right)^2 \leq \frac{1}{\pi^2} \int_0^1 (\partial_x u)^2{\rm d}x$$
in Daniel's notes the choice is $\phi(s) = s\log(s)$ with $\psi(s) = 2s^{\frac 12}$, leading to the well-known log Sobolev inequality, here we just need to take $\phi(s) = -\log(s)$ with $\psi(s) = \log(s)$, and the game is over.
Remark: I definitely welcome any other different approaches!
