Suppose we have a strictly increasing function $f$ such that $f'(x) \geq 2xf(x)$ under the interval $(0,1)$. Suppose we have a strictly increasing function $f: [0, 1] \to \mathbb{R}$ such that $f'(x) \geq xf(x)+x$ under the interval $(0,1)$. It is known that $f(0)=0$. How would one find the minimum value of $f(1)$, given that $f(x)$ is continuous in $[0,1]$ and differentiable in $(0,1)$?
I tried using the AM-GM inequality to get that $f'(x) \geq 2f(x)$ and the fact that the integral of $\frac{f'(x)}{f(x)}$ is $\ln(f(x)) + c$, but I didn't get anywhere.
 A: By the Mean Value Theorem applied to the function $x \mapsto \arctan f(x) - \frac{x^2}{2}$ on $[0, 1]$, it follows that there exists $\xi \in (0, 1)$ satisfying
$$ \arctan f(1) - \frac{1}{2}
= \frac{f'(\xi)}{1+f(\xi)^2} - \xi
\geq 0. $$
So $f(1) \geq \tan(1/2)$ for any such function $f$. Moreover, this lower bound is achieved by the function $f(x) = \tan(x^2/2)$. Therefore the answer is $\tan(1/2)$.
A: The idea is that to minimize $f$ we must minimize $f'$ at every point, i.e. make the inequality into an equality. Solving that would use separation of variables:
$$ \frac{\mathrm{d}y}{y^2+1}=x\,\mathrm{d}x \quad \implies \quad \tan^{-1}(y)=\frac{x^2}{2} $$
(The constant of integration is $0$ since $f(0)=0$.)
To make this rigorous, note the inequality is equivalent to
$$ \frac{\mathrm{d}}{\mathrm{d}x}\left[\tan^{-1}(y)\right]\ge x. $$
Inequalities may be (definitely) integrated, so apply $\int_0^1 \mathrm{d}x$ to both sides to get
$$ \tan^{-1}(y)\ge \frac{x^2}{2}. $$
Thus $f(x)\ge\tan(x^2/2)$, and $f(1)\ge\tan(1/2)$.
A: Observe,
$$  \frac{f'(x)}{1+f(x)^2} = [\tan^{-1}f(x)]'\geq x $$
Integrate both sides over $x \in (0,1)$ with starting value $f(0) = 0$,
$$\tan^{-1}f(1)\ge\frac12$$
Thus,
$$f(x)\ge \tan\frac12$$
