This question has already been asked:
Proving Inequality using Induction $a^n-b^n \leq na^{n-1}(a-b)$
However has not been answered properly. (Even thought the OP checked an answer)
The answers provided are direct proofs and not induction proofs.
Question: Suppose $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then:
$$a^n-b^n \leq na^{n-1}(a-b)$$
Anybody has ideas?
Best Answer
We do the induction step. Suppose that for a certain $k$ we have $a^k-b^k\le ka^{k-1}(a-b)$. We will show that $a^{k+1}-b^{k+1}\le (k+1)a^k(b-a)$.
We have $$a^{k+1}-b^{k+1}=a^{k+1}-a^kb+a^kb-b^{k+1}=a^k(a-b) +b(a^k-b^k).\tag{1}$$ Note that by the induction hypothesis $b(a^k-b^k)\le a(a^k-b^k)\le ka^k(a-b)$. So the sum on the right-hand side of (1) is $\le (k+1)a^k(a-b)$.