show that $h$ is a differentiable and $h(-1)=h(0)=h'(0)=0$ and $h(1)=1$ then $h^{(3)}\geq 3$ If $h$ is an real function, differentiable three times on $[-1,1]$, such that $h(-1)=h(0)=h'(0)=0$ and $h(1)=1$. Prove that exist an real number $r\in(-1,1)$ such that $h^{(3)}(r)\geq 3$

For this I used Taylor's theorem so
\begin{eqnarray}
h(x)= h(a)+h'(a)(x-a)+\frac{h''(a)}{2!}(x-a)^{2}+\frac{h^{(3)}(a)}{3!}(x-a)^{3}
\end{eqnarray}
for $a\in[-1,1]$, then, for $a=0$
\begin{eqnarray}
h(x)= \frac{h''(0)}{2!}x^{2}+\frac{h^{(3)}(0)}{3!}x^{3}
\end{eqnarray}
and for $a=1$, and $a=-1$
\begin{eqnarray}
h(x)= 1+h'(1)(x-1)+\frac{h''(1)}{2!}(x-1)^{2}+\frac{h^{(3)}(1)}{3!}(x-1)^{3}\\
h(x)=h'(-1)(x+1)+\frac{h''(-1)}{2!}(x+1)^{2}+\frac{h^{(3)}(-1)}{3!}(x+1)^{3}
\end{eqnarray}
but I don't the way to continue. Do you know some hint to continue?

We can solve this by considering the function $h$ over the intervals $(0,1)$ and $(-1, 0)$.  Taylor's theorem tells us that there is a point $x \in (0, 1)$ such that:
$$h(1) = h(0) + h'(0) + h''(0)/2 + h^{(3)}(x)/6.$$
Similarly, we also know there is a point $y \in (-1, 0)$ such that:
$$h(-1) = h(0) -h'(0) + h''(0)/2 - h^{(3)}(y)/6.$$ Then from the known values of $h$ we find that $h^{(3)}(x) + h^{(3)}(y) = 6$. So either $h^{(3)}(x) \geq 3$ or $h^{(3)}(y) \geq 3$.
