Proving $f'''(x)≥3$ exists where $f(-1)=0$, $f(0)=0$, $f(1)=1$, $f'(0)=0$ Let $f:[-1,1]\to\mathbb{R}$ be $C^3$ with $f(-1)=0$, $f(0)=0$, $f(1)=1$, and $f'(0)=0$. I need to show that there exists a point $x\in(-1,1)$ such that $f'''(x)≥3$. I would like some feedback on my solution. I am also curious if it's possible to use concavity from supposing $g'''<0$ to find another proof.
First, we apply MVT thrice to find points $-1<a_1<a_2<a_3<0$ such that
$$f'(a_1)=f(0)-f(-1)=0=\frac{f'(0)-f'(a_1)}{-a_1}=f''(a_2) $$
and
$$f'''(a_3)=\frac{f''(0)-f''(a_2)}{-a_2}=\frac{f''(0)}{-a_2}.$$
Now let $g(x)=f(x)-\frac{1}{2}x^2(x+1)$ so $g''(x)=f''(x)-3x-1$. Repeating the same process as above for $g$, we find points $0<b_2<b_1<1$ such that
$$g'(b_1)=g(1)-g(0)=0 \implies g''(b_2)=g'(b_1)-g'(0)=0.$$
Suppose now for the sake of contradiction that $f''''(x)<3$ for all $x\in(-1,1)$. Then $f''(0)<-3a_2$ and by MVT, there is some point in $(a_2,b_2)$ for which $g''(b_2)-g''(a_2)<3(b_2-a_2)$ so $-(-f''(a_2)+3a_2+1)>3(b_2-a_2)>3a_2$ and thus $a_2<-\frac{1}{6}$. It follows that $f''(0)<\frac{1}{2}$, which means $g''(0)=f''(0)-1<0$. Therefore, by MVT, there is a point $c\in(0,b_2)\subset(-1,1)$ such that
$$g'''(c)=f'''(c)-3=\frac{g''(b_2)-g''(0)}{b_2}>0\implies f'''(c)>3.$$

From Taylor we know that
$$f(x) = \frac12f''(x)x^2+\frac16f'''(\xi)x^3,$$
hence
$$
\begin{align}
0=f(-1)&=\frac12f''(0)\cdot(-1)^2+\frac16f'''(\xi_1)\cdot (-1)^3\\
1=f(1)&=\frac12f''(0)\cdot1^2+\frac16f'''(\xi_2)\cdot 1^3.
\end{align}
$$
Subtracting gives
$$6=f'''(\xi_1)+f'''(\xi_2)$$
which would be impossible if $f'''(x)<3$ for all $x\in[-1,1]$.
