If $f\geq0 , f(0)=f(1)=0$,and $\int_0^1|f''|/f dx$ exists,how to prove $\int_0^1f''/f dx\geq \pi^2$ If $f\in C^2 , f\geq0$ on $[0,1]$ , $f(0)=f(1)=0$,and $\int_0^1f''/f dx$ exists,how to prove  $$\int_0^1\frac{|f''|}{f}dx\geq \pi^2$$
The lower bound $\pi^2$ was guessed by myself, if it was not true ,then how to find this lower bound ?
My way is to have an analytic  extension with $T=2$, and use the Fourier Series to have an evaluation .
But it seems like that I was wrong.
 A: Let:
$$
f(x) = \sin(\pi x)+0.1\sin(2 \pi x)
$$
Then 
$$
f'(x) = \pi\cos(\pi x)+0.2\pi\cos(2 \pi x)
$$
$$
f''(x) =-\pi^2\sin(\pi x)-0.4\pi^2\sin(2 \pi x)
$$
So:
$$
\int_0^1 \frac{|f''(x)|}{f(x)} dx=\int_0^1  \frac{\pi^2\sin(\pi x)+0.4\pi^2\sin(2 \pi x)}{\sin(\pi x)+0.1\sin(2 \pi x)}dx = 9.25905... < \pi^2 = 9.8696
$$
Here's the link to the result:
https://www.wolframalpha.com/input/?i=%5Cint_0%5E1++%5Cfrac%7B%5Cpi%5E2%5Csin(%5Cpi+x)%2B0.4%5Cpi%5E2%5Csin(2+%5Cpi+x)%7D%7B%5Csin(%5Cpi+x)%2B0.1%5Csin(2+%5Cpi+x)%7Ddx
Indeed it is hard to come up with a non-sine function that allows the integral to converge, but if you make a composition of such kind, you can likely lower your bound. I would suspect however, that there should be a sequence of functions $f_n$ such that the integral converges to zero. 
Edit
My reasoning to claim that the integral can be made arbitrarily close to zero is that with a large enough sine series, it should be possible to build $f(x)$ such that the integral exists due to the characteristics of the sine on the extremes, but outside the extremes and some mid point, the function is "flat", i.e. with arbitrarily small  $f''(x)$ but with large values for $f(x)$, hence decreasing the value of the integral. 
A: The proposition you're trying to prove is false. This integral can be made arbitrarily close to $4$. A family of examples:
$$f(x) = \begin{cases}2x&0\le x\le \frac12-\epsilon\\ 1-\frac{3\epsilon}{4} - \frac{3}{2\epsilon}(x-\frac12)^2 + \frac1{4\epsilon^3}(x-\frac12)^4  & \frac12-\epsilon\le x\le \frac12+\epsilon\\ 2-2x&\frac12+\epsilon\le x\le 1\end{cases}$$
Why is this $C^2$? We check the values and derivatives where the formulas change. At $\frac12+\epsilon$ using the middle formula,
\begin{align*}f\left(\frac12+\epsilon\right) &= 1-\frac{3\epsilon}{4} - \frac{3}{2\epsilon}\epsilon^2 + \frac1{4\epsilon^3}\epsilon^4 = 1 +\epsilon\left(-\frac34-\frac32+\frac14\right) = 1-2\epsilon\\
f'\left(\frac12+\epsilon\right) &= -\frac{3}{\epsilon}\epsilon+\frac1{\epsilon^3}\epsilon^3 = -2\\
f''\left(\frac12+\epsilon\right) &= -\frac{3}{\epsilon} + \frac{3}{\epsilon^3}\epsilon^2 = 0\end{align*}
Those match the values from the other formula it meets, and the first two derivatives are continuous there. At $\frac12-\epsilon$, we appeal to symmetry; $f(x)=f(1-x)$, so continuity of the first two derivatives at $\frac12+\epsilon$ implies it at $\frac12-\epsilon$.
What is $\int_0^1 \frac{|f''(x)|}{f(x)}\,dx$? Rather than calculate it exactly, I'll estimate. We have
$$f''(x) = \frac3{\epsilon}\left(-1+\left(\frac{x-\frac12}{\epsilon}\right)^2\right) \le 0$$
for $\frac12-\epsilon\le x\le\frac12+\epsilon$, so $\int_{1/2-\epsilon}^{1/2+\epsilon}|f''(x)|\,dx = \left|f'\left(\frac12+\epsilon\right)-f'\left(\frac12-\epsilon\right)\right| = 4$. Now, we divide by $f(x)$ inside the integral; from $1-2\epsilon\le f(x)\le 1-\frac34\epsilon$ on the interval,
$$\frac{4}{1-\frac34\epsilon}\le\int_{1/2-\epsilon}^{1/2+\epsilon}\frac{|f''(x)|}{f(x)}\,dx \le\frac{4}{1-2\epsilon}$$
What about $f$ outside that subinterval? The second derivative out there is identically zero, so that part contributes nothing to the integral, and $\int_0^1 \frac{|f''(x)|}{f(x)}\,dx$ lies between those bounds.
As $\epsilon\to 0$, both of those bounds tend to $4$, and therefore the integral does by the squeeze theorem. The limiting function isn't $C^2$, but we can make the integral arbitrarily close to $4$ this way.
