Inequality – Proving Inequality Using Induction

Question

I was trying to prove this inequality using induction, but couldn't do.

Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:

$$a^n-b^n \leq na^{n-1}(a-b)$$

Answer 1

You will want to use that $$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}$$

What can you say about the powers of $a,b$ given $0<b<a$?*

SPOILER

Since $0<b<a$, we have $0<b^k<a^k$

thus

$$\begin{align} a^n-b^n&=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}\\&<(a-b)\sum_{k=0}^{n-1}a^{n-k-1}a^{k}\\&=(a-b)\sum_{k=0}^{n-1}a^{n-1}\\&=(a-b)na^{n-1}\end{align}$$