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
$$g'''(c)=f'''(c)-3=\frac{g''(b_2)-g''(0)}{b_2}>0\implies f'''(c)>3.$$

This isn't true. Consider the function:
$$f(x) = \frac{1}{2} x(x+1)$$
Observe that $f(-1) = 0$, $f(0) = 0$ and $f(1) = 1$. On the other hand, $f'''(x) = 0$ for all $x \in (-1,1)$.
Edit:
Hmm there seems to be a problem somewhere in the middle. You write:
$$g''(b_2)-g''(a_2) = -g''(a_2) < 3(b_2-a_2)$$
But then you write:
$$-(-g''(a_2))> 3(b_2-a_2) > 3a_2$$
right after that and I'm not entirely sure how this follows. See, we could have that $a_2 = -\frac{1}{2}$, $b_2 = \frac{1}{2}$. Then, $3(b_2-a_2) = 3$ and $-(3a_2+1) = \frac{1}{2}$.
