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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

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Why is it valid to integrate both sides of that inequality? –  Ethan Jan 24 '13 at 7:01
    
@Ethan: If you integrate from $1$ to $x$, you get the result [in reply to deleted comment]. Update: It is valid because the functions are integrable (continuous even) and $g(x)\leq h(x)$ for all $x$ implies $\int_a^b g(x)dx\leq \int_a^bh(x)dx$ for all $a<b$. –  Jonas Meyer Jan 24 '13 at 7:04

3 Answers 3

up vote 6 down vote accepted

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 $x\geqslant1$.

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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 $$\sqrt{f(x)}\geq\sqrt{f(1)}+\frac{1}{2}(x-1)$$

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Fully detailed proof, ready to be copied and handed back? Hmmm... –  Did Jan 24 '13 at 7:21

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.

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