Quick but not simple question. $2^\sqrt2$ or e, which is greater? 
$2^\sqrt2$ vs $e$, which is greater?

$(2^\sqrt2)^\sqrt2 = 4\quad $  &  $\quad e^\sqrt2$ = ?
$\log(2^\sqrt2) = \sqrt2\log(2)\quad$ & $\quad \log(e) = 1$
I tried but can't induce comparable form.
Is anybody know how to prove it?
 A: This is the same as comparing $\frac{3}{2}\log(2)$ and $1$. Since $x(1-x)$ is non-negative and bounded by $\frac{1}{4}$ on $(0,1)$, we have
$$ 0\leq\int_{0}^{1}\frac{x^2(1-x)^2}{1+x}\,dx \leq \frac{1}{16}$$
where the middle integral is exactly $-\frac{11}{4}+4\log(2)$. It follows that 
$$ \frac{33}{32} \leq \frac{3}{2}\log(2) \leq \frac{135}{128} $$
so $\frac{3}{2}\log(2)>1$ and $\color{red}{2\sqrt{2}>e}$. This proof just requires a polynomial division, perfectly doable by hand.
About $\sqrt{2}\log(2)$, we have
$$ \log(2)=\lim_{n\to +\infty}\sum_{k=n+1}^{2n}\frac{1}{k}\leq\lim_{n\to+\infty}\sum_{k=n+1}^{2n}\frac{1}{\sqrt{k}\sqrt{k-1}}\stackrel{\text{CS}}{\leq}\lim_{n\to +\infty}\sqrt{n\sum_{k=n+1}^{2n}\left(\frac{1}{k-1}-\frac{1}{k}\right)}$$
and the RHS is exactly $\frac{1}{\sqrt{2}}$. This is just a slick application of creative telescoping and the Cauchy-Schwarz inequality.
A: Instead of comparing $2^{\sqrt 2}$ and $e$, let's raise both to $\sqrt 2$ and compare $2^2$ and $e^{\sqrt 2}$:
$$
e^{\sqrt 2} > 2.7^{1.4} \approx 4.017068799 > 4 = 2^2
$$
Or use that
$$
e^x > 1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}
$$
with $x=1.41$ and get
$$
e^{\sqrt 2} > e^{1.41} > 4.03594 > 4
$$
In fact, $e^{\sqrt 2} \approx 4.113250377 > 4$.
A: If you know that $\ln(2)\approx0.69$ and $1/\sqrt2=\sqrt2/2\approx1.414/2=0.707$, then you have $\ln(2)\lt1/\sqrt2$, in which case $\ln(2^\sqrt2)=\sqrt2\ln2\lt1=\ln(e)$, hence $2^\sqrt2\lt e$.
It's not hard to show that $\sqrt2\gt1.4$, since $1.4^2=1.96\lt2$.  It's a little trickier to show that $\ln(2)\lt0.7$, but this can be done by comparing the area beneath the curve $y=1/x$ to the areas of the trapezoids containing it with endpoints at $x=1$, $4/3$, $5/3$, and $2$:
$$\ln(2)=\int_1^2{dx\over x}\lt{1\over6}\left(1+2\cdot{3\over4}+2\cdot{3\over5}+{1\over2} \right)={1\over6}\left(1+{3\over2}+{6\over5}+{1\over2} \right)={1\over6}\cdot{42\over10}={7\over10}$$
A: $2\sqrt{2}^2 = 8$
$e^2 < 2.8*2.8 = 7.84$
You're welcome
A: Let $f(x)=\ln(x),\,g(x)=x^{-1/2}$. By the Taylor series of $e^x$, we have
$$e^{-0.35}>1-0.35+\frac12\left(0.35\right)^2-\frac16\left(0.35\right)^3=0.704>0.7$$
Hence, $g(e^{0.7})-f(e^{0.7})=e^{-0.35}-0.7>0$. Applying Taylor series again shows
$$e^{0.7}>1+0.7+\frac{1}{2}(0.7)^2+\frac{1}{6}(0.7)^3=2.002>2$$
Observe that $f$ and $g$ are respectively strictly increasing and decreasing over $(0,\infty)$, so $g>f$ holds for all $x$ in $(0,2.002)$. Therefore $g(2)>f(2)$, which rearranges to $e>2^{\sqrt{2}}$. These calculations are perfectly feasible to do by hand.
