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

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 $\xi < a$.
