TITLE: Bounding $f'$ in terms of $f$ and $f''$
QUESTION [7 upvotes]: Assume that $f: \mathbb{R} \to [0,\infty)$ is $C^2$ and $|f''(x)| \leq A$ for all $x$.  Show that the inequality 
$$(f'(x))^2 \le 2Af(x)$$ holds for all $x$.
The hint given in the question was, "Taylor's theorem."
I had thought in the beginning that the bound was pretty straightforward, but I've been stuck for a bit now.  
I have that, centering my Taylor expansion about some point $a$, gives 
$$f(x) = f(a) + f'(a)(x-a) + \frac{f^{''}(\psi)}{2!}(x-a)^2$$
for some $\psi$ in the interval $(x,a)$.  The last term is the Lagrange remainder.
Then $$f(x) = f(a) + f'(a)(x-a) + \frac{f^{''}(\psi)}{2!}(x-a)^2$$
$$\le f(a) + f'(a)(x-a) + \frac{|f^{''}(\psi)|}{2!}(x-a)^2$$
$$\le f(a) + f'(a)(x-a) + \frac{A}{2!}(x-a)^2$$
$$\implies f(x)-f(a)\le  f'(a)(x-a) + \frac{A}{2!}(x-a)^2$$
$$\implies \frac{f(x)-f(a)}{x-a}\le  f'(a) + \frac{A}{2!}(x-a)$$
$$\implies f'(\eta)\le  f'(a) + \frac{A}{2!}(x-a)$$
(applying the Mean Value Theorem on the left hand side)
and this is where I am stuck.  
Any hints are welcome.
Thanks,

REPLY [4 votes]: You have $\frac{A}{2}(x-a)^2+f'(a)\cdot(x-a)+f(a)\geq f(x)\geq 0$ for all $x,a\in\mathbb{R}$.  Hence, the quadratic $\frac{A}{2}t^2+f'(a)\cdot t+f(a)$ takes only nonnegative values for $t\in\mathbb{R}$.  As $A\geq 0$, the discriminant of this quadratic is nonpositive:
$$\big(f'(a)\big)^2-4\left(\frac{A}{2}\right)f(a)\leq 0\,.$$
Since $a$ is arbitrary, you get your desired result.