Prove that there exists $\xi \in (a,b)$ such that $f(a)-2f(\frac{a+b}{2})+f(b)=\frac{1}{4}(b-a)^2f''(\xi) .$ 
Given that f is twice differentiable on $[a,b]$, prove that there exists $\xi \in (a,b)$ such that
$$f(a)-2f\left(\frac{a+b}{2}\right)+f(b)=\frac{1}{4}(b-a)^2f''(\xi) .$$

This problem was given in a book and the hint was to consider the following auxiliary function, and apply Rolle's Theorem:

$$F(t) = f(t) +f(a) -2f\left(\frac{t+a}{2}\right) - \lambda\cdot\frac{(t-a)^2}{4}$$
Where $$\lambda = \frac{f(a)+f(b)-2f\left(\displaystyle\frac{a+b}{2}\right)}{\displaystyle\frac{(b-a)^2}{4}}.$$

I tried and got stuck.
My attempt:
Notice that $F(a) =  F(b) = 0$, by Rolle's Theorem, there exists $\xi_1 \in (a,b)$ such that $F'(\xi_1)=0$.
After differentiation, we obtain $$F'(t) = f'(t)-f'\left(\displaystyle\frac{t+a}{2}\right)-\lambda\cdot\displaystyle\frac{t-a}{2}$$
We find that $F'(a) = 0$, thus applying Rolle's Theorem again, there exists $\xi_2 \in (a,\xi_1)$ such that $F''(\xi_2) = 0$
Therefore
$$0=f''(\xi_2)-\frac{1}{2}f''\left(\frac{\xi_2+a}{2}\right) - \frac{\lambda}{2}$$
and
$$\lambda = 2f''(\xi_2) - f''\left(\frac{\xi_2+a}{2}\right).$$
I got stuck here and was unable to proceed. I did consider using Darboux's Theorem but I don't think that it is applicable here.
 A: Your proof is almost complete at the stage when $F'(\xi_1)=0$ ie $$f'(\xi_1)-f'((\xi_1+a)/2)-\lambda\cdot\frac{\xi_1-a}{2}=0$$ Just use the mean value theorem to write $$f'(\xi_1)-f'((\xi_1+a)/2))=\frac{\xi_1-a}{2}f''(\xi)$$ for some $\xi\in(a, b) $ and then you get $\lambda=f'' (\xi) $.
There is no need to differentiate again and deal with $F''(t) $.
A: I can only suggest a different approach. Consider $g(\xi) = f(\xi) - \frac{\lambda \xi^2}{2}$, where $\lambda$ is the same as in your attempt. Then $f''(\xi) = \lambda$ is equivalent to $g''(\xi) = 0$. If we assume the converse then $g''$ is positive or is negative on the whole interval $[a,b]$ because $g''$ is continuous. This means that $g''$ is a convex or a concave function. On the other hand, we have
$$
g(a) - 2g\left(\frac{a + b}{2}\right) + g(b) = f(a) - 2f\left(\frac{a + b}{2}\right) + f(b) - \frac{\lambda (a^2  - \frac{(a + b)^2}{2}+ b^2)}{2} =
f(a) - 2f\left(\frac{a + b}{2}\right) + f(b) - \frac{\lambda (a - b)^2}{4} = 0. 
$$
Which contradicts the geometric interpretation of convexity/concavity of $g$.
