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?
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 $\le (k+1)a^k(a-b)$.
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 ≤ 1*a^0(a−b) = a − b,
so the statement holds.
(2) Second base case (n = 2):
a^2 − b^2 = (a + b)(a − b)
< 2a(a − b) since a > b
= na^(n−1)(a − b), for n = 2
(3) Inductive step: Now suppose that a^n − b^n < na^(n−1)(a − b), where 0 <
b < a. Then
a^(n+1) − b^(n+1)= a^(n+1)- a^n*b + a^n*b − b^(n+1) = a^n(a − b) + b(an^ − b^n)
< a^n(a − b) + b^n*a^(n−1)(a − b) by the inductive hypothesis
< a^n(a − b) + ana^(n−1)(a − b) since b < a
= (n + 1)a^n(a − b)
so that a^(n+1) − b^(n+1) < (n + 1)a^n(a − b) for n > 2.
This image depicts my own, fairly easy to follow, solution. I found it by trying to recreate the original form by manipulating the k+1 expression.
