How prove $\pi^2>2^\pi$

Question

show that $$\pi^2>2^\pi$$ I use computer found $$\pi^2-2^\pi\approx 1.044\cdots,$$

can see this

I know $$\Longleftrightarrow \dfrac{\ln{\pi}}{\pi}>\dfrac{\ln{2}}{2}$$ so let $$f(x)=\dfrac{\ln{x}}{x}$$ so $$f'(x)=\dfrac{1-\ln{x}}{x^2}=0,x=e$$ so $f(x)$ is Strictly increasing on $(2,e)$, and is Strictly decreasing on $(e,3) $ so I can't know $f(2)$ and $f(\pi)$ which is bigger?

maybe this problem exsit have easy methods by hand

Answer 1

Hint: $$\frac{\ln 2}{2}=\frac{\ln 4}{4}.$$

Answer 2

Actually, we can prove $9 > 2^{\pi}$.

We are going to prove $\dfrac{9}{8} > 2^{0.16}$, i.e $\left(\dfrac{9}{8}\right)^6 > 2^{0.96}$

To see this we prove $\left(\dfrac{9}{8}\right)^6 > 2$, which can be verified by a bit direct computation

Answer 3

This is similar to this problem, asking which of $\pi^3$ and $3^\pi$ is bigger. It can be deduced from the inequalities

$$3\lt\pi\lt{22\over7}$$

Specifically,

$$2^{11}=2048\lt2187=3^7\implies2^{22/7}\lt3^2\implies 2^\pi\lt\pi^2$$

Answer 4

For $n>e$ $f(x)$-decreasing so $ln(\pi)/\pi>ln(4)/4$ but $ln(4)/4=ln(2)/2$