<s>[INST] Which number is greater, $2^\sqrt2$ or $e$? 
Claim: $\color{red}{2^\sqrt2<e}$

Note: $2^\sqrt2=e^{\sqrt2\ln2}$
Different approach: We show $e^{x-1}>x^\sqrt x$ for $x>2$.
Let $f(x) = x -1 - \sqrt{x} \ln x $. We have $f'(x) = 1 - \frac{ \ln x }{2 \sqrt{x} } - \frac{1}{\sqrt{x}}$. By inspection, note that $f'(1)=0$ and since for $x>1$, pick $x=4$, for instance, we have $f'(4)=1- \frac{ \ln 4 }{4} - \frac{1}{2} = \frac{1}{2} - \frac{ \ln 2 }2 > 2$, then we know $f(x)$ is increasing for $x> 1$, Thus 
$$ f(x) > f(1) \implies x - 1 - \sqrt{x} \ln x > 0 \implies x - 1 \geq \sqrt{x} \ln  x \implies e^{x-1} > x^{\sqrt{x}}  $$
Putting $x=2$, we obtain
$$e>2^\sqrt2$$
What do you think about this approach? Is there an easier way?
 [/INST] To show $2^{\sqrt{2}}<e$, it suffices to show that $\sqrt{2}\ln 2 < 1$, or $\ln 2 < \frac{1}{\sqrt{2}}$. Notice that $\ln 2 = \int\limits_{1}^{2}{\frac{1}{x}\,dx}$, and by Cauchy-Schwarz we have
$$\ln 2 = \int\limits_{1}^{2}{\frac{1}{x}\cdot 1\,dx}\le\left(\int\limits_{1}^{2}{\frac{1}{x^2}\,dx}\right)^{1/2}\left(\int\limits_{1}^{2}{1\,dx}\right)^{1/2} = \left(\frac{1}{2}\right)^{1/2}\left(1\right)^{1/2} = \frac{1}{\sqrt{2}}.$$
Equality cannot hold as $(1/x)/1$ is not constant, so it follows that $\ln 2 < \frac{1}{\sqrt{2}}$, as desired.
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