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Let $f:[0,1] \to \mathbb{R}$ be $C^2$ such that $f(0) = f(1) = 0$ and $f''(x) \geq -1$. Given these conditions, what is the maximum possible value (or a tight upper bound) of $f$?

Intuitively, I have the idea.

If $f(c) = M$ is the maximum, $M$ cannot be too large, because $f'(c) = 0$ and $f'$ must quickly decrease (by the mean value theorem) to $$f'(\xi) = \frac{M}{c-1} $$ and if $M$ is quite large this value is a large (negative) number. Therefore, since $f''$ is bounded below, $M$ must also be bounded.

I'm not sure how to formalize it though.

Note: I would appreciate a solution (or a hint, etc.) which does not utilize Taylor's theorem.

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    I don't think you can find the exact maximum value. However you can find an upper bound which works for all $f$. – 2017-01-29
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    @FrankLu Okay, in that case replace "maximum" with "tight upper bound". – 2017-01-29
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    Perhaps you could make an argument that the $f$ with the greatest maximum will be the one satisfying $f''(x)=1$ i.e. $f(x)=-\frac 12 x(x-1)$ – 2017-01-29

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Fix the function $$g(x) = \frac{x - x^2}{2}$$ as in WW1's comment, with $g'' \equiv -1.$ If $f$ is any other function satisfying the conditions then $f-g$ is convex with zeros at $0$ and $1$, and therefore $$(f-g)(t) = (f-g)\Big((1-t) \cdot 0 + t \cdot 1\Big) \le (1-t) \cdot (f-g)(0) + t \cdot (f-g)(1) = 0,$$ i.e. $f(t) \le g(t)$ everywhere. So the maximum will be the maximum of $g$, attained at $g(1/2) = 1/8.$