For a twice differentiable function defined on $(a,\infty)$, prove a given inequality involving sup of $|f(x)|,|f'(x)|$ and $|f''(x)|$

Question

The question:

Let $f$ be a twice differentiable real-valued function defined on $(a,\infty)$, where $a\in \mathbb R$. Let $M_i=\sup |f^{(i)}(x)|, i=0,1,2$. Then prove that $M_1^2\le 4M_0 M_2$.

My solution attempt:

Let $c\in (a,\infty)$, then by Taylor formula on $[c,x]$ we have, $$f(x)=f(c)+f'(c)(x-c)+\frac {f''(c)}2 (x-c)^2+o((x-c)^2), x\to c, x>c$$ or, $$\frac{f''(c)}2 (x-c)^2+f'(c)(x-c)+f(c)=f(x)+o((x-c)^2)\implies |\frac{f''(c)}2| (x-c)^2+|f'(c)|(x-c)+|f(c)|\ge 0\tag 1$$ $(1)$ being a quadratic on LHS, holds necessarily if $ |f'(c)|^2\le 2 |f(c)||f''(c)|\le 2M_0 M_2\le 4 M_0 M_2\implies |f'(c)|\le \sqrt {4M_0M_2}\implies {4M_0M_2}$ is an upper bound for $|f'(c)|$ and so $\sup |f'(c)|\le {4M_0M_2}\implies M_1^2\le4M_0M_2$.

Is that correct? Thanks.