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.

Hint: It helps to write your inequality as $\frac 1e >\frac {\ln \pi}{\pi}$ and to note that $\frac 1e = \frac {\ln e}e$. Think about $f(x)=\frac {\ln x}x$. — lulu, Mar 26 '18 at 11:27
This is $e^\pi>\pi^e$, an old chestnut, which I am sure has been tackled here several times. — Macavity, Mar 26 '18 at 12:11
@Macavity Funny enough, searching MSE for old chestnut returns Comparing $\pi^e$ and $e^\pi$ without calculating them as the second hit ;-) — dxiv, Mar 28 '18 at 22:25

score 2 · Answer 1 · answered Mar 26 '18 at 11:24

2

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

answered Mar 26 '18 at 11:24

Dr. Sonnhard Graubner

95,283

score 0 · Accepted Answer · edited Mar 28 '18 at 22:20

0

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 $e$: $$\large{e^{\frac1e}>\pi^\frac1\pi}$$

edited Mar 28 '18 at 22:20

answered Mar 26 '18 at 12:29

user

154,566

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

2 Answers2