Request theorem/Proof for $F(x)>0$ for $a<x<b$

Suppose $$F(a)=0, F(b)=0, a<b$$ and $F''(x)<0$ for $a<x<b$.

Do we have $F(x) > 0$ for $a<x<b$?

Can you tell me an existing theorem that I can use in a paper to obtain the conclusion? I think there should be one, but I can't remember.

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If $F(x) < 0$ for some $a < x < b$, then you can use the mean value theorem three times to show that $F''(X) > 0$ somewhere. –  arkeet Dec 11 '12 at 7:51
@arkeet I mean, can you tell me an existing theorem that I can use in a paper to obtain the conclusion? I think there should be one, but I can't remember. –  Chen Dec 11 '12 at 8:14

This is a consequence of the convexity of $-F(x)$ on $[a,b]$.

Suppose $F(c) < 0$ for some $c \in (a,b)$. By the mean value theorem, there exist values $d_1 \in (a,c)$ and $d_2 \in (c,b)$ such that \begin{align} F'(d_1) = \frac{F(c) - F(a)}{c - a} = \frac{F(c)}{b-a} < 0 \\ ~ \\ F'(d_2) = \frac{F(b) - F(c)}{b - c} = \frac{-F(c)}{c-b}> 0, \end{align} and hence $F'(x)$ changes sign somewhere on $(a,b)$. This contradicts $F''(x) < 0$. Hence, $F(c) \geq 0$ for all $c \in (a,b)$.

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How about this proof: By Rolle's theorem, $F'(x) = 0$ at least once in the interval $(a, b)$. However it must happen exactly once since $F''(x) < 0$... This critical point is a maximum point, so we have $F(x)$ ascending from $(a, 0)$ towards a maximum and then descending back to $(b, 0)$.