Show inequality (probably using the induction)

Question

I'm having a problem with this inequality.

Show that for every integer k $\ge 2$ and n:

$\frac{n^{k+1}}{k+1} \ge 1^k + 2^k +...+(n-1)^k $

Answer 1

It is a consequence of Bernoulli's inequality:

For all $n\in\mathbf N$, if $h>-1$, then $\enspace(1+h)^n\ge 1+nh$.

Indeed, suppose, by the inductive hypothesis, $\;\dfrac{n^{k+1}}{k+1} \ge 1^k + 2^k +...+(n-1)^k\;$ for some $n\ge 0$. We deduce that $$\frac{n^{k+1}}{k+1}+n^k \ge 1^k + 2^k +...+(n-1)^k+n^k,$$ so to prove the inductive step, it will be enough to prove that \begin{align} \frac{(n+1)^{k+1}}{k+1}\ge\frac{n^{k+1}}{k+1}+n^k&\iff(n+1)^{k+1}\ge n^{k+1}+(k+1)n^k \\% &\iff \biggl(1+\frac1{n}\biggr)^{k+1}\ge1+(k+1)\frac 1n, \end{align} which is exactly Bernoulli's inequality.

Answer 2

A different approach: Notice

$$ \frac{n^{k+1}}{k+1}= \int\limits_0^n x^k dx \geq \sum_{i=0}^{n-1} \frac{n}{n} \left( i \frac{n}{n} \right)^k = 1^k + 2^k + ... + (n-1)^k $$