If $f''(x)+f(x)>0$ and $f(x)>0$ $\forall x\in(a,b)$; $f(a)=f(b)=0$; prove that $b-a>\pi$. Please help me to solve this question:

Suppose $f:[a,b] \to \Bbb R$ satisfies:

*

*$f''(x)+f(x)>0$ and $f(x)>0$ for all $x\in(a ,b)$;

*$f(a)=f(b)=0$.

Prove that $b-a>\pi$.

Thanks in advance.

I think I have a solution when $f''$ is assumed to be continuous. 
When $x\in (a,b)$, we have that $\sin\left(\frac{x-a}{b-a}\pi\right)>0$ so using the first assumption,
$$0<\int_a^bf(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx+\int_a^bf''(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx=I_1+I_2\tag{1}.$$
We have, integrating by parts and using $f(a)=f(b)=0$ that 
$$I_1=-\frac{\pi}{b-a}\left(\int_a^b-f'(x)\cos\left(\frac{x-a}{b-a}\pi\right)dx.\right)$$
Doing the same for $I_2$, we finally get 
$$\small 0<\left(\frac{(b-a)^2}{\pi^2}-1\right)\int_a^bf'(x)\cos\left(\frac{x-a}{b-a}\pi\right)=\left(\frac{(b-a)^2}{\pi^2}-1\right)\int_a^bf(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx.$$
As $\int_a^bf(x)\sin\left(\frac{x-a}{b-a}\pi\right)dx>0$, we are done.
