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Really need a small hint for this one. Suppose $f$ is a function such tha $f(x)>0$ and $f'(x)$ is continuous at every x. If $f'(t)\ge \sqrt{f(t)}\forall t$ then show that $\sqrt{f(x)}\ge\sqrt{f(1)} +\frac{(x-1)}{2}$

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You can use the Taylor series of $f(x)$ around $x=1$:

$$f(t) = f(1) + f'(1)(x-1)+R(x) > f(1) +\sqrt{f(1)}(x-1)$$

Then because $\sqrt{}$ is monotonic, and again using a Taylor expansion:

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

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    $\begingroup$ This is wrong. You can't know that $R(x)>0$, and the final inequality is backwards. $\endgroup$ Apr 18 '12 at 6:38
  • $\begingroup$ @HaraldHanche-Olsen is right - the solution above is indeed wrong. Instead, you can take the derivative of $\sqrt{f(t)}$ to find: $f'>\sqrt{f}$, and then use the Taylor expansion of $\sqrt{f}$. $\endgroup$
    – nbubis
    Apr 18 '12 at 11:13
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Try differentiating $\sqrt{f(t)}$, perhaps?

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Consider g(t) defined as $\sqrt{f(t)}$. $g'(t) = \frac{f'(t)}{2\sqrt{f(t)}} \ge \frac{1}{2} \forall t$ by the condition given. Hence $g(x) - g(1) \ge \frac{x-1}{2}$ as if not then there would exist t between 1 and x where $g'(t) \lt \frac{1}{2}$.

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