# Prove that $a^n-b^n \geq (a-b)nb^{n-1}$ where $a \geq b \geq 0$ [duplicate]

This question already has an answer here:

I am having trouble with this proof. I believe the way to do it is through induction. This is what I have so far.

Proof:

We begin by induction on n. For the case that n = 1, we have $a^1-b^1= (a-b) \geq (a-b)(1)(b^0) = (a-b)$. $(a-b) \geq (a-b)$

Now we assume that this is true for some natural number k. $a^k-b^k \geq (a-b)kb^{k-1}$

Now we must show it is true for k + 1. So $a^{k+1} - b^{k+1} = a^ka-b^kb.$

I am not really sure how to proceed from this point. Where can I use the inductive hypothesis?

## marked as duplicate by Martin R, haqnatural, heropup, C. Falcon, Claude LeiboviciFeb 11 '17 at 6:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Do you have to use induction? – Niklas Feb 7 '17 at 22:28
• You should state the conditions on $a,b,n$. Presumably, $n$ is a positive integer, and a,b are real numbers such that $a \ge b$. Yes? If so, you should edit the problem to make that clear. – quasi Feb 7 '17 at 22:32
• @quasi, $a\ge b$ is not required here. – Wolfram Feb 7 '17 at 22:33
• @Wolfram: I think some restriction such as the one I suggested is required. For example, if $n = 2$, $a = 1$, $b = 2$, then the inequality fails. – quasi Feb 7 '17 at 22:39
• @quasi It is $a\geq b$ (in the title) :) – Niklas Feb 7 '17 at 22:40

## 4 Answers

I'm not sure that induction is the best way to go here. What I would use is the following property:

$$\frac{a^n-b^n}{a-b} = a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}$$

With induction, it's rather difficult to get the factor of $n$ working properly on the RHS (although I suspect it's possible).

• Darn, beat me to it :D – Simply Beautiful Art Feb 7 '17 at 22:29
• @SimplyBeautifulArt me too XD – Brevan Ellefsen Feb 7 '17 at 22:29
• @BrevanEllefsen Glad to know my idea is the same as others'. – Carl Schildkraut Feb 7 '17 at 22:30

Consider $f\colon \mathbf R \to \mathbf R$ with $f(x)=x^n$. Assume $0\leq b\leq a$. By mean value theorem $$a^n -b^n=f(a)- f(b) = f'(\xi) (a-b) \geq nb^{n-1}(a-b)$$ where $\xi \in (b,a)$.

Another way which came into my mind is the following which is using integration:

$$a^n -b^n = \int_b^a nx^{n-1} \; \mathrm d x \geq nb^{n-1}\int_b^a 1 \; \mathrm d x = nb^{n-1}(a-b).$$

• Nice use of calculus here! – Carl Schildkraut Feb 7 '17 at 22:36
• Thank you. This is always the first idea that comes to my mind when I work with inequalites of this type. – Niklas Feb 7 '17 at 22:37
• @Niklas. What you are using is the mean value theorem, not the intermediate one :) – ThePuix Feb 7 '17 at 22:41
• @ThePuix Thank you, of course. Switching from german to english and reverse is sometimes tricky :P – Niklas Feb 7 '17 at 22:43
• This version has the advantage that it works for nonnegative real values of $n$, not just nonnegative integer values. – quasi Feb 7 '17 at 22:52

You can proceed by induction if you wish (though admittedly it is not as elegant as the other solutions). From assuming

$$a^k - b^k \geq (a-b) kb^{k-1}$$

this can be clearly rearranged to give

$$a^k \geq (a-b) kb^{k-1} + b^k.$$

Then, for $a\geq 0$:

\begin{align} a^{k+1} - b^{k+1} &= a\cdot a^k - b^{k+1} \\ & \geq a(a-b) kb^{k-1}+ab^k - b^{k+1} \\ & = (a-b) \left(\frac{a}{b}k+1\right) b^k \end{align}

where the last equality follows after a bit of algebra. Then, if $a\geq b> 0$ (i.e. $\frac{a}{b}\geq 1$) we have $\frac{a}{b}k+1\geq k+1$ and the inductive step follows.

Obviously this inductive proof explicitly requires $a\geq b> 0$.

Trivial case $b=0$ aside, let $c = \frac{a}{b} \ge 1$ then, after dividing by $b^n \gt 0 \,$, the inequality becomes:

$$c^n-1 \geq (c-1)n$$

Let $d = c-1 \ge 0\,$, then, after rearranging, the inequality can be written as:

$$(1+d)^n \geq 1 + n d$$

The latter is just Bernoulli's inequality and, since both previous steps were reversible equivalences, this proves the originally proposed inequality.

• Nice reasoning. – marty cohen Feb 8 '17 at 6:55