Show that a definite integral of $f(x)$ from $-1$ to $1$ is greater than or equal to $2f(0)$

Here's a problem I need some help with:

Let $f$ be a twice differentiable function in the closed interval $[-1, 1]$ and $f''(x) \geq 0$ for all $x \in [-1, 1]$. Show that $$\int^1 _{-1} f(x)dx \geq 2f(0)$$

When does the equality hold?

There was a small hint given to apply the Mean Value Theorem and the fact that $f'(x)$ is growing in that interval.

I've got a very vague idea how to apply the MVT. This is what I've done so far $$\int_{-1} ^1 f(x) = F(1) - F(-1) = (1 -(-1))f(\xi) = 2f(\xi)$$ using the MVT. But I've got no idea how to show the inequality to be true and why $f(0)$ in particular. Actually I'm not sure if I'm on the right track to begin with.

PS. I'd prefer some hints to a complete solution at first.

• Is it $f'(0)$ in the inequality or $f(0)$? – Alamos Apr 29 '15 at 23:02
• $f(x)=x^2$ says it can't be $f'(0)$. – Chappers Apr 29 '15 at 23:03
• You can consider the line $y=f'(0)x+f(0)$. The condition on the derivative tells you the graph of $f$ is above that line. Compare areas under the graphs. – Alamos Apr 29 '15 at 23:09

Taylor's theorem with the integral form of the remainder gives $$f(x) = f(0) + x f'(0) + \int_0^x (x-t) f''(t) \, dt$$ (Nothing fancy here—just some integration by parts.)
The last integral is nonnegative for all $x$ by the condition on $f''$. Hence $f(x) \geqslant f(0)+xf'(0)$. Now, as Alamos notes, you integrate this inequality over $[-1,1]$.
Hint: Integrate by parts twice \begin{align} \int_{-1}^1f(x)\,\mathrm{d}x &=\int_{-1}^0f(x)\,\mathrm{d}x+\int_0^1f(x)\,\mathrm{d}x\\ &=f(0)-\int_{-1}^0(x+1)f'(x)\,\mathrm{d}x+f(0)-\int_0^1(x-1)f'(x)\,\mathrm{d}x\\ &=2f(0)-\frac12f'(0)+\int_{-1}^0\frac{(x+1)^2}2f''(x)\,\mathrm{d}x\\ &+\frac12f'(0)+\int_0^1\frac{(x-1)^2}2f''(x)\,\mathrm{d}x\\ &=2f(0)+\int_{-1}^0\frac{(x+1)^2}2f''(x)\,\mathrm{d}x+\int_0^1\frac{(x-1)^2}2f''(x)\,\mathrm{d}x \end{align}