How to prove that $\pi > e\ln(\pi)$?

Question

How can you prove the following inequality? $$\pi > e\ln(\pi)$$

Edit: I have tried the following, $\pi = e^{\ln(\pi)}$ so: $$e^{\ln(\pi)} > e\ln(\pi)$$ $$e^{\ln(\pi)-1} > \ln(\pi)$$ $$\ln(\pi)-1>\ln(\ln(\pi))$$

I am unsure where to go from here.

Answer 1

Note that $\sqrt[x] x$ is a decreasing function for $x\ge e$ then since $\pi>e$

$$\large{\pi > e\ln \pi}$$ $$\large{\frac1e>\frac{\ln \pi}\pi}$$ $$\large{\frac1e > \ln(\pi^{\frac1\pi})}$$ Putting both as an exponent of $e$: $$\large{e^{\frac1e}>\pi^\frac1\pi}$$

Answer 2

HINT: define the function $$f(x)=x-e\ln(x)$$ for $$x>3$$ and use calculus,