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 $0 < b < a$. Prove that if $n$ is a positive integer, then:

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

Solution 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

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

Solution 2:

You may use the mean value theorem to show this: Define $f(x) = x^n$ on $[b, a]$, clearly, $f(x)$ is differentiable in $(b, a)$ and continuous on $[b, a]$. By MVT, there exists $\xi \in (b, a)$ such that $$a^n - b^n = f(a) - f(b) = f'(\xi)(a - b) = n\xi^{n - 1}(a - b) \leq na^{n - 1}(a - b)$$ as $\xi < a$.