If $f(x)$ is 2x differentiable in $(a,b)$ & $f'(a)=f'(b)=0$, prove that, $\exists\xi $ in $(a,b)$ S.T. $|f''(\xi )|\leq\frac{4(f(b)-f(a))}{(b-a)^{2}}$ Here is my argument (it doesn't feel 100% correct for some reason):
By the mean value theorem, there exists $\xi_{1}$ in $(a,b)$ such that,
$$f'(\xi_{1}) =  \frac{f(b)-f(a)}{b-a}$$
Since, $f'(a)=f'(b)=0$, then by the mean value theorem again, there exists $\xi_{2}$ in $(\xi_{1},b)$ such that,
$$f''(\xi_{2})=\frac{f'(b)-f'(\xi_{1})}{b-\xi_{1}}=\frac{-(f(b) -f(a))}{(b-a)(b-\xi_{1})}$$
Since $\xi_{1}$ can be no lower than $a$,
$$f''(\xi_{2}) \geq \frac{-(f(b) -f(a))}{(b-a)^{2}} \, or\,  f''(\xi_{2}) \leq \frac{f(b) -f(a)}{(b-a)^{2}} \leq \frac{4(f(b) -f(a))}{(b-a)^{2}}$$
$$Assuming\, f(b)\geq f(a)$$
So it follows,
 $|f''(\xi_{2} )| \leq  \frac{4(f(b)-f(a))}{(b-a)^{2}}$?

As $f'(b) = f'(a)=0$, there is $y\in (a, b)$ such that $f''(y)=0$.
