How do I compare $\sqrt{2}$ and $\pi^{1/ \pi}$?

Here is a plan of attack.

Study the function $f(x)=x^{1/x}$.

  • Show that $f'(x)<0$ when $x>e$.
  • Compare $f(4)$ and $f(\pi)$.

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 $2^{11}=2048 < 2187 = 3^7$.