How to show $\sqrt{f(x)} \geq \sqrt{f(1)} +\frac{1}{2}(x-1)$?

Question

Suppose $f$ is a function such that $f(x) > 0$ and $f'(x)$ is continuous at every real number $x$. If $f'(t) \geq \sqrt{f(t)}$ for all $t$, then show that $$\sqrt{f(x)} \geq \sqrt{f(1)} +\frac{1}{2}(x-1)$$ for all $x \geq 1$.

Trial: Here $f'(t) \geq \sqrt{f(t)} \implies f(t) \geq \int \sqrt{f(t)}~ dt$ Then I am unable to solve

Answer 1

Hint: Let $g=2\sqrt{f}$. Then $g$ is well defined since $f\gt0$ everywhere, and $g'\geqslant1$. Hence $g(x)\geqslant g(1)+x-1$ for every $x\geqslant1$.

Answer 2

Using your first inequality $$f'(t) \geq \sqrt{f(t)}$$ Divide both sides by $\sqrt{f(t)}$ $$f'(t)\frac{1}{\sqrt{f(t)}}\geq 1$$ Divide both sides by 2, $$f'(t)\frac{1}{2\sqrt{f(t)}}\geq \frac{1}{2}$$ Note $\frac{d}{dt} \sqrt{f(t)}=f'(t)\frac{1}{2\sqrt{f(t)}}$, so integrating both sides gives.. $$\int_{1}^xf'(t)\frac{1}{2\sqrt{f(t)}} \ dt \geq \int_{1}^x \frac{1}{2} \ dt $$ Which equals $$\sqrt{f(x)}-\sqrt{f(1)}\geq \frac{1}{2}(x-1)$$ And thus $$\sqrt{f(x)}\geq\sqrt{f(1)}+\frac{1}{2}(x-1)$$

Answer 3

Let $h(x)=\sqrt{f(x)}- \sqrt{f(1)} -\frac{1}{2}(x-1)$, and then $$h'(x)=\frac{f'(x)}{2\sqrt{f(x)}} -\frac{1}{2}\ge0$$ because $f'(x) \geq \sqrt{f(x)}$. Since $h(1)=0$, we're done.