How do I show this :$\pi^n>n^{\pi} $ for $n\neq 3$?

Question

le $n$ be a positive integer such that $n\neq 3$ really i have tried to show this inequality $\pi^n>n^{\pi} $ that is true for all positive integer n different from $ n=3 $

Attempt: from the titled inequality for $n>3$ i have : $$n \log \pi >\pi \log n \tag {1}$$ . from $(1)$ we have: $\displaystyle\frac{n}{\pi}> \displaystyle\frac{\log n}{\log \pi}\tag{2}$ , but the latter say nothing to us then my question is:

Question:How do I show this :$\pi^n>n^{\pi} $ for $n\neq 3$ if it is true ?

Answer 1

HINT

Consider the function $f(x) = \pi\ln x - x \ln \pi$ for a fixed $n$ and find extrema using ordinary Calculus.

Answer 2

HINT: your inequality its equivalent to this other $\frac{n}{\ln n}>\frac{\pi}{\ln \pi}$ for $n\neq 3$. Thus it is enough to study the function $f(x):=\frac{x}{\ln x}$.

$f'(x)=0\iff \ln x=1$

Answer 3

Your computations are fine. Another way of thinking about this is to see that the statement that you want to prove is equivalent to$$\frac n{\log n}>\frac\pi{\log\pi}.$$Now, consider the map $x\mapsto\dfrac x{\log x}$ and prove that it attains its minimal value when $x=e$ (which is smaller than $3$). Since $\frac4{\log4}\simeq2.89$ and $\frac\pi{\log\pi}\simeq2.74$, all you have to do is to check that the inequality holds when $n=1$ and when $n=2$.