prove $m^{m-1} < (m-1)^m$ for m > 3

Question

I found that if m > 3 then $m^{m-1} < (m-1)^m$ for m > 3 seems to hold true for a lot of cases. Can someone prove this inductively ?

Answer 1

1) When $m=4$, then $(m-1)^m=3^4>4^3=m^{m-1}$, so the inequality holds for $m=4$.

2) Assume that $(m-1)^m>m^{m-1}$ for some integer $m\ge4$.

Then $\displaystyle m^{m+1}=\frac{m^{m+1}}{(m-1)^m}\cdot(m-1)^m>\frac{m^{m+1}}{(m-1)^m}\cdot m^{m-1}=\frac{(m^2)^m}{(m-1)^m}>\frac{(m^2-1)^m}{(m-1)^m}$

$\;\;\;\;\;\;\displaystyle=\frac{(m-1)^{m}(m+1)^m}{(m-1)^m}=(m+1)^m$, so the assertion holds for $m+1$.

Therefore $(m-1)^m>m^{m-1}$ for all integers $m\ge4$.

Answer 2

Take logarithms. We just have to prove that $$ f(n)=\frac{n}{\log n} $$ is an increasing function for $n\geq 3$, or that $$ g(n) = \frac{\log n}{n} $$ is a decreasing function on the same set.

This is a one-shot-one-kill problem through derivatives, since: $$ g'(x)=\frac{1-\log x}{x^2} $$ gives that $g$ is a decreasing function on the set $x>e$.

Answer 3

You can prove that $f\colon x\in (1,\infty)\to x\ln(x-1) - (x-1)\ln x$ is strictly positive for $x > 2$. It is increasing, and cancels at $2$.