Bounding $f'$ in terms of $f$ and $f''$ 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,
 A: 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.
A: Here, a very important detail is that the function is non-negative. If the value $f(x)$ is too low and $f'(x)$ has a large absolute value, then since the growth rate is bounded in terms of A, the function $f$ will necessarily slide into the negative region. First, you have to show that for any $a,x\in[0,\infty)$
\begin{equation}
f(x)\leq f(a) + f'(a) (x-a) + \frac{A}{2}(x-a)^2=g_a(x),
\end{equation}
what you have already shown.
Assume that $A>0$. Without loss of generality $a=0$, otherwise make a shift $x\mapsto x+a$. Now, note that the parabola $g_0$ takes it's minimum in $x_0=-\frac{f'(0)}{A}$. The value of $g_0$ in $x_0$ is
\begin{equation}g_0(x_0)=f(0)-\frac{f'(0)^2}{A}+\frac{f'(0)^2}{2A},
\end{equation}
which is zero iff $f(0)=\frac{f'(0)^2}{2A}$, and negative iff $f'(0)^2>2A f(0)$. Since $f$ maps into $[0,\infty)$, we necessarily need $f'(0)^2\leq 2A f(0)$.
As I mentioned, with the shift $\tilde f_a(x)=f(a+x)$, we now get that $\tilde f\ '_a(0)^2\leq 2A \tilde f_a(0)$, thus $f'(a)^2\leq 2A f(a)$ for any $a\in \mathbb{R}$.
The case $A=0$ is trivial, we have a linear function $f$, which is non-negative if $f'(x)$ is zero.
