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I want to prove a simple theorem about contour integration via residues and I need the following estimation:
$e^{a\sqrt{r}} > r$ for any real a > 0 and r >> 0.

Is this true? If so, what is an easy way of seeing this?

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    Presumably, you know the following inequality: $e^x\ge 1+x$. You could work from there.2011-07-09
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    Expand the exponential into a power series, observe that all summands are positive. Can you prove that $C r^{\alpha} \gt r$ for $C \gt 0$ and $r \gg 0$ provided $\alpha \gt 1$? @Raskolnikov: that's not good enough because of the square root.2011-07-09
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    Wouldn't Raskolnikovs hint work if I substitute $(1/2)a\sqrt{r}$ for x, use the inequality and then square both sides? Then I obtain ${a\sqrt{r}}+(1/4)a^{2}r$ on the RHS which I can estimate against r for r >> 0.2011-07-09
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    Well, careful, squaring both sides of an inequality isn't guaranteed to work because both sides could be in (0,1). It works here because the exponential is greater than 1 when its argument is positive.2011-07-09
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    Yes it would because of anon's comment. Eli why don't you work your solution into an answer. I'll take a look afterwards. Yes, the inequality is usually proved using the power series expansion2011-07-09
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    The inequality holds for every nonnegative $r$ if and only if $a\ge2/\mathrm{e}$. The inequality cannot hold uniformly over every positive $a$ for any fixed positive $r$, and not even uniformly for $r\ge|\log a|^2/a^2$ (but $r\ge|\log a|^{2+\varepsilon}/a^2$ will do).2011-07-09

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Unless the "r>>0" of the current post is a typo, it may be intended to mean that $r$ is much bigger than $0$.

If that is the case, then for any fixed $a$, the inequality holds if $r$ is large enough. Note for example that if we use the ordinary power series for $e^x$ we find that $$e^{a\sqrt{r}} >1+ar^{1/2}+\frac{a^2r}{2!}+\frac{a^3r^{3/2}}{3!}.$$

The term $a^3r^{3/2}/3!$ is by itself larger than $r$ if $r$ is big enough.

Added: Or else for simplicity let $r=x^2$. We want to compare $e^{ax}$ with $x^2$. A L'Hospital's Rule calculation shows that $$\lim_{x\to\infty}\frac{x^2}{e^{ax}}=0.$$

One disadvantage of this approach is that we get no explicit bounds from the calculation.

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    "r>>0" wasn't a typo, I should've been more precise with what I meant. Thanks though for giving me a solution.2011-07-09
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    @Eli: I thought it likely wasn't a typo, the language sounded like that of someone who knew what he/she was doing.2011-07-09