How do I compare $\sqrt{2}$ and $\pi^{1/ \pi}$?
I applied in calculator, I got $\pi^{1/ \pi}=1.4396194958475907$ and $\sqrt{2}=1.414213562373095$. So, $\pi^{1/ \pi} > \sqrt{2}$. How to show analytically? I could expand the Taylor series expansion of $\sqrt{1+x}=1+\frac{x}{2}-\frac{1}{8}x^2+...$ But what will I do with $\pi^{1/ \pi}$. Please help me.
Here is a plan of attack.
Study the function $f(x)=x^{1/x}$.
Notice that you can write
$$\sqrt{2}=2^{1/2}=4^{1/4}$$
So you are just comparing
$$4^{1/4},\pi^{1/\pi}$$ But that's the same function. Investigate $$f(x)=x^{1/x}$$ and find where it's increasing and where it's decreasing, and its behaviour in general.
EDIT: $2$ and $\pi$ are on the opposite sides of a maximum (which is at $e$) so you'll need to transform one of your values a little. @JyrkiLahtonen has a brilliant insight that $2^{1/2}=4^{1/4}$. Without that, I was searching for much more complicated transformations that didn't end up useful at all.
We just have to compare $2^\pi$ and $\pi^2$. It is well known that $3<\pi<\frac{22}{7}$, and
$$ 2^\pi < 2^{22/7} \color{red}{<} 3^2 < \pi^2 $$ is a straightforward consequence of $2^{11}=2048 < 2187 = 3^7$.
$$\pi^{1/\pi}>\sqrt 2\iff\pi^{2/\pi}>2\iff\pi^2>2^\pi$$ So our goal is now to see if $2^\pi$ is greater or lesser than $\pi^2$.
For that, consider the function $$f(x)=\ln\frac{2^x}{x^2}=x\ln 2-2\ln x$$ Its derivative is $$f'(x)=\ln 2-\frac2x$$ which vanishes at $c=2/\ln 2$.
But $$f(c)=2-2\ln\left(\frac2{\ln 2}\right)=2-2\ln2+2\ln\ln 2$$ which is positive since $\ln 2<1$.
Since $\lim_{x\to0}f(x)=\lim_{x\to\infty}f(x)=\infty$, $f$ has a minimum at $c$, so $f$ is positive.
What we need is that $f(\pi)>0$, which implies $2^\pi>\pi^2$.
Let $$ f(x)=\frac{\log(x)}x $$ then $$ f'(x)=\frac{1-\log(x)}{x^2} $$ Thus, $f(x)$, and therefore $x^{1/x}=e^{f(x)}$, is decreasing when $x\gt e$. Since $e\lt\pi\lt4$, we have $$ e^{1/e}\gt\pi^{1/\pi}\gt4^{1/4}=2^{1/2} $$
