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 $\le (k+1)a^k(a-b)$.

  • 1
    $\begingroup$ I am often awed by the mathematical wit the top members of this community have. Vous êtes un maître monsieur André Nicolas. $\endgroup$ – ex.nihil Mar 6 '16 at 21:34

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.

enter image description here

  • $\begingroup$ It's really worth a good look. Wraps up nicely at the end $\endgroup$ – zkunning Jun 24 '19 at 1:02

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