# $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then $a^n - b^n \leq na^{n-1}(a - b)$.

I'm taking a basic discrete math course and I'm having a hard time with Mathematical Induction.

The problem is stated as:

Suppose that $a$ and $b$ are real numbers with $0 < b < a$. Prove that if $n$ is a positive integer, then $a^n - b^n \leq na^{n-1}(a - b)$.

### The work I have so far is:

• Basis Step: \begin{aligned} P(1) \equiv a^1 - b^1 &\leq (1)a^{(0)}(a - b) \\\ a - b &\leq (1)(1)(a - b) \\\ a - b &\leq a - b \end{aligned}

• Inductive Step: Assume that $P(k)$ is true for any fixed integer $k \geq 0$. Induction Hypothesis: $$P(k) \equiv a^k-b^k \leq ka^{k-1}(a-b)$$ Prove that $P(k+1)$ is true given $P(k)$: \begin{aligned} P(k+1) &\equiv& a^{k+1} - b^{k+1} &\leq (k+1)a^k(a-b)\\\ &\equiv& a\cdot a^k - b \cdot b^k &\leq (k+1)a^k(a-b)\\\ &\equiv& a\cdot a^k - b \cdot b^k &\leq ka^k(a-b) + a^k(a-b)\\\ &\equiv& a\cdot a^k - b \cdot b^k &\leq a \cdot \underbrace{\left( ka^{k-1}(a-b) \right)}_{IH} + a^k(a-b)\\\ \end{aligned}

### Where I'm At

I don't really know where to go from there. I think I'm supposed to play around with the left side such that I can find the other part of my Induction Hypothesis on the left side, but I'm lost at that point.

• I don't think $0$ counts as a positive integer, so you should refactor your base case for $n = 1$. – Waleed Khan Mar 14 '13 at 3:47
• True, I forgot about that somewhere in my transcription from paper to computer. – AntoineG Mar 14 '13 at 3:47

$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})\le (a-b)(a^{n-1}+a^{n-1}+\ldots$$
• For the right part of the inequation, how does $(a - b)(a^{n-1} + a^{n-1} + \ldots$ go past the dots? – AntoineG Mar 14 '13 at 14:39
Since $b<a$, then $0<a-b$, so $a^n-b^n\le na^{n-1}(a-b)$ if and only if $$\frac{a^n-b^n}{a-b}\le na^{n-1}.\tag{1}$$ We can write the left hand side of $(1)$ as $$\sum_{k=0}^{n-1}a^kb^{n-1-j}.\tag{2}$$ Use the fact that $0<b<a$ to get the rest of the way. There is no need for induction if you know that $(2)$ holds.