Suppose a ∈ R1, f is a twice-differentiable real function on (a,∞), and M0, M1, M2 are the least upper bounds of $|f(x)|, |f'(x)|, |f''(x)|,$ respectively, on $(a, ∞)$.
Prove that $M_1^2 ≤ 4M_0M_2$.
Hint: If $h > 0,$ Taylor’s theorem shows that $f'(x) = \frac{1}{2h} [f(x + 2h) − f(x)] − hf''(ξ)$ for some $ξ ∈ (x, x + 2h). $ . . .
My question is how to apply Taylor's theorem so that we get $$f'(x) = \frac{1}{2h} [f(x + 2h) − f(x)] − hf''(ξ)$$ for some $ξ ∈ (x, x + 2h).$
For any $x$ and $y$, $f(y) = f(x) + f'(x)(y-x) + \frac{1}{2} f''(\xi)(y-x)^2,$ is the Taylor approximation where $\xi$ is some point in between $x$ and $y$.
Thus,
$$f'(x)(y-x) = f(y) - f(x) - \frac{1}{2}f''(\xi)(y-x)^2.$$
If $|f(x)| \leqslant M_0$ and $|f''(x)| \leqslant M_2$ are bounds, then
$$|f'(x)| \leqslant \frac{|f(x)| + |f(y)|}{|y-x|} + \frac{1}{2} |f''(\xi)| |y-x| \leqslant \frac{2M_0}{|x-y|}+ \frac{M_2}{2}|x-y|.$$
Now notice that
$$\min_{0 < z < \infty} \left(\frac{2M_0}{z} + \frac{M_2z}{2}\right)= 2 \sqrt{M_0M_2}$$
You should be able to finish from here.