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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?

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    Do you have to use induction? – 2017-02-07
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    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. – 2017-02-07
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    @quasi, $a\ge b$ is not required here. – 2017-02-07
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    @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. – 2017-02-07
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    @quasi It is $a\geq b$ (in the title) :) – 2017-02-07
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    @Niklas Hebestreit: Was that always there? If so, I must have missed it. – 2017-02-07
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    @quasi I think it got added in the end. So dont worry ;) – 2017-02-07
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    @quasi $-3\ge-4$ for your example. Non-negativity is essential, I guess. – 2017-02-07

4 Answers 4

1

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.

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    Nice reasoning. – 2017-02-08
8

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:

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

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    Nice use of calculus here! – 2017-02-07
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    Thank you. This is always the first idea that comes to my mind when I work with inequalites of this type. – 2017-02-07
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    @Niklas. What you are using is the mean value theorem, not the intermediate one :) – 2017-02-07
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    @ThePuix Thank you, of course. Switching from german to english and reverse is sometimes tricky :P – 2017-02-07
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    This version has the advantage that it works for nonnegative real values of $n$, not just nonnegative integer values. – 2017-02-07
6

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).

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    Darn, beat me to it :D – 2017-02-07
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    @SimplyBeautifulArt me too XD – 2017-02-07
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    @BrevanEllefsen Glad to know my idea is the same as others'. – 2017-02-07
3

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 $a\geq b> 0$.