If $f'(t)≥\sqrt{f(t)}$ then prove that $\sqrt{f(x)}≥\sqrt{f(1)}+\frac{1}2(x-1)$

Question

QUESTION: Suppose $f$ is a function such that $f(x)>0$ and $f'(x)$ is continuous for every real number $x$. If $f'(t)≥\sqrt{f(t)}$ then prove that $\sqrt{f(x)}≥\sqrt{f(1)}+\frac{1}2(x-1)$.

MY APPROACH: We observe that graph lies entirely above the $x$-axis. And I had assumed that $f(x)=kx^{2n}$ $(k>0)$. That satisfies the given conditions.. but that's not a general proof. How may I prove it rigorously?

Any help is much appreciated. Thank you so much..

Answer 1

Notice that $$\left(\sqrt{f}\right)'(x) = \frac{f'(x)}{2\sqrt{f(x)}} \ge \frac12$$ so by the mean value theorem for every $x > 1$ there exists some $\theta \in \langle 1, x\rangle$ such that $$\sqrt{f(x)} - \sqrt{f(1)} = \left(\sqrt{f}\right)'(\theta) \underbrace{(x-1)}_{\ge0} \ge \frac12(x-1)$$ which proves the claim for $x > 1$.

On the other hand, for $x \in \langle 0,1]$ the statement is trivial since $$\sqrt{f(x)} \ge \sqrt{f(1)} \ge \sqrt{f(1)} + \underbrace{\frac12(x-1)}_{\le 0}$$ and the first inequality holds since $\left(\sqrt{f}\right)' \ge \frac12 \ge 0$ so $\sqrt{f}$ is increasing on $\langle 0, +\infty\rangle$.

Answer 2

Hint: the assumption implies that $(\sqrt{f(t)})'\ge \frac 12$. Integrate.

Answer 3

Hint: $$\frac{\mathrm d}{\mathrm dt}\sqrt{f(t)}=\frac{1}{2\sqrt{f(t)}}f'(t).$$