Prove that $a^n-b^n \geq (a-b)nb^{n-1}$ where $a \geq b \geq 0$ 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?
 A: 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:
$$a^n -b^n = \int_b^a nx^{n-1} \; \mathrm d x \geq nb^{n-1}\int_b^a 1 \; \mathrm d x = nb^{n-1}(a-b).$$
A: 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). 
A: 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 $a\geq b> 0$. 
A: 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.
