Show that $f:(0,\infty) \to (0,\infty)$ differentiable and such that $f'(x) \le \frac{1}{2f(x)}$ implies $f(x) \le \sqrt{x}$ Problem: let $f:(0,\infty) \to (0,\infty)$ be a differentiable function such that
$$\lim_{x \to 0^+} f(x)=0 \\ f'(x) \le \frac{1}{2f(x)}, \ \forall x>0$$
Show that $f(x) \le \sqrt{x}$ for any $x>0$.
My textbook solves this problem using the derivative of a product, but I've tried this other approach: since $f$ has codomain $(0,\infty)$, it is $f(x)>0$ for any $x>0$ and so, being by hypothesis $f'(x) \le \frac{1}{2f(x)}$ for any $x>0$, it is
$$f'(x) \le \frac{1}{2f(x)} \iff f(x)f'(x) \le \frac{1}{2}$$
By hypothesis $f$ is differentiable in $(0,\infty)$, hence it is continuous in $(0,\infty)$ and so, for $a>0$, it is integrable in any interval $[a,x] \subset (0,\infty)$. So
$$f(x)f'(x) \le \frac{1}{2}, \ \forall x>0 \implies \int_a^x f(t)f'(t)dt\le\int_a^x \frac{dt}{2}$$
$$\iff f(x)^2 \le x+f(a)^2-a$$
Letting $a \to 0^+$ in the inequality and using the hypothesis that $f(x) \to 0$ when $x \to 0^+$, it follows that $f(x)^2 \le x$. But $f(x)>0$ and the domain of $f$ is $(0,\infty)$, hence $x>0$ as well and so the inequality $f(x)^2 \le x$ is equivalent to the one obtained taking the square root both sides and hence $f(x) \le \sqrt{x}$. Is this correct? In particular, I am not quite sure about the fact that I integrated in a subset $[a,x]$ and then let $a \to 0^+$.
 A: Once you have $f(x)^2 \le x+f(a)^2-a$ for $0 < a < x$ you can indeed take the limit for $a \to 0^+$ and conclude that $f(x)^2 \le x$. That part is fine.
But there is still a problem with your approach: You are assuming that $f(x)f'(x)$ is integrable, which is not given. What you are using is the fundamental theorem of calculus for $g(x) = f(x)^2$:
$$
 g(x) - g(a) = \int_a^x g'(t) \, dt \, .
$$
This is true if $g'(x) = 2f(x) f'(x)$ is Riemann integrable, or more generally, if $g'$ is Lebesgue integrable (which is equivalent to $g$ being absolutely continuous on compact intervals).
A sufficient condition is (for example) that $f'$ is continuous.
This additional condition is avoided if you argue instead that $f(x)^2-x$ is decreasing because its derivative is $\le 0$ (which might be what your textbook solution does).
Roughly speaking: $g' \le 0$ implies that $g$ is decreasing because of the mean-value theorem, and that requires only the differentiability of $g$. Using the fundamental theorem of calculus for the same conclusion requires stronger conditions on the given function.
