Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$ I was trying to prove this inequality using induction, but couldn't do.

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

$$a^n-b^n \leq na^{n-1}(a-b)$$
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 $0&lt;b&lt;a$?*

SPOILER


   Since $0&lt;b&lt;a$, we have $0&lt;b^k&lt;a^k$


thus 


  $$\begin{align} a^n-b^n&amp;=(a-b)\sum_{k=0}^{n-1}a^{n-k-1}b^{k}\\&amp;&lt;(a-b)\sum_{k=0}^{n-1}a^{n-k-1}a^{k}\\&amp;=(a-b)\sum_{k=0}^{n-1}a^{n-1}\\&amp;=(a-b)na^{n-1}\end{align}$$
The proof proceeds by induction. Here we will need to make use of two base
cases, since it turns out that $n = 1$ needs to be treated separately.

(1) First base case $(n = 1)$: here $a^1 −b^1 \le 1 \times a^0(a−b) = a − b$,

so the statement holds.

(2) Second base case $(n = 2)$:

$a^2 − b^2 = (a + b)(a − b)$

$&lt; 2a(a − b)$    since $a &gt; b$

$= na^{n−1}(a − b)$, for $n = 2$

(3) Inductive step: Now suppose that $a^n − b^n &lt; na^{n−1}(a − b)$, where $0&lt;  b &lt; a$. Then

$a^{n−1} − b^(n+1)= a^{n−1}- a^n \times b + a^n \times b − b^(n+1) = a^n(a − b) + b(a^n − b^n)$

$&lt; a^n(a − b) + b^n \times a^{n−1}(a − b)$ by the inductive hypothesis

$&lt; a^n(a − b) + ana^{n−1}(a − b)$ since $b &lt; a$

$= (n + 1)a^n(a − b)$

so that $a^{n−1} − b^{n−1} &lt; (n + 1)a^n(a − b)$ for $n &gt; 2$.