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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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    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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    @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