I am stuck with this problem.
$f : [a,b] \to \mathbb R$ is double differentiable on $[a,b]$ and suppose $|f''(x)|\leq M ~ \forall x \in [a,b]$. Let $c \in (a,b)$. Then $|f(c)-f(a)-(c-a)\frac{f(b)-f(a)}{b-a}|\leq \frac{1}{8} M(b-a)^2$
I can prove this when the upper bound is $M(b-a)^2$.
$|f(c)-f(a)-(c-a)\frac{f(b)-f(a)}{b-a}|=|c-a| |\frac{f(c)-f(a)}{c-a}-\frac{f(b)-f(a)}{b-a}|$. This equals $|c-a||\frac{f'(\xi_1)-f'(\xi_2)}{\xi_1-\xi_2}||\xi_1-\xi_2|$ for some $\xi_1\in (a,c)$ and $\xi_2 \in (a,b)$. This equals $|c-a||f''(\psi)||\xi_1-\xi_2|\leq (b-a)^2M$
Although I know this is weak and I haven't exploited the $\xi '$s fully. I cannot find the solution afterer lots of tries, so any help is appreciated.
Hint: The polynomial $$ p(x)=f(a)+(x-a)\frac{f(b)-f(a)}{b-a} $$ interpolates $(a,f(a))$ and $(b,f(b))$. What can you say about $$ f(x)-p(x)-(f(\xi)-p(\xi))\cdot\frac{(x-a)(x-b)}{(\xi-a)(\xi-b)}? $$