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.

My answer is under the assumption that $f$ and $f'$ are continuous on $[a,b]$ and $f''$ exists on $(a,b)$. 
Define
$$g:[a,b]\to \mathbb{R}, \quad x\mapsto f'(x)\sin(x-a)-f(x)\cos(x-a).$$
By definition, $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Moreover, since $f(a)=f(b)=0$, $g(a)=0$ and $g(b)=f'(b)\sin(b-a)$. Then by mean value theorem, there exists $c\in(a,b)$, such that
$$\frac{f'(b)\sin(b-a)}{b-a}=\frac{g(b)-g(a)}{b-a}=g'(c)=(f''(c)+f(c))\sin(c-a).\tag{1}$$
On the one hand, beacause $f(x)>0$ on $(a,b)$ and $f(b)=0$, $f'(b)\le 0$; on the other hand, $f''(c)+f(c)>0$. Combing these facts with $(1)$, we can conclude that $b-a>\pi$.
Edit: After reading Sanchez's answer, I realized that without assuming that $f$ is differentiable at $a,b$, the proof can be modified as follows.
Assume that $b-a\le \pi$. Then $g$ is differentiable on $(a,b)$ and 
$$g'(x)=(f''(x)+f(x))\sin(x-a)>0.$$ 
Therefore, $g$ is strictly increasing on $(a,b)$. As a result, both 
$$g(a^+):=\lim_{x\to a^+} g(x)=\lim_{x\to a^+}f'(x)\sin(x-a)$$
and
$$g(b^-):=\lim_{x\to b^-} g(x)=\lim_{x\to b^-}f'(x)\sin(x-a)$$
exist, and $g(b^-)>g(a^+)$.
However, since $f(x)>0$ on $(a,b)$ and $f(a)=f(b)=0$, by mean value theorem, 
$$\limsup_{x\to a^+}f'(x)\ge\liminf_{x\to a^+}\frac{f(x)-f(a)}{x-a}\ge 0 $$
and 
$$\liminf_{x\to b^-}f'(x)\le\limsup_{x\to b^-}\frac{f(x)-f(b)}{x-b}\le 0 .$$
It follows that $g(a^+)\ge 0\ge g(b^-)$, a contradiction.
