Estimating formula for $a^n-b^n$

Question

Let $a,b$ be real numbers, with $0< a< b$. From the identity $$b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+\ldots +a^{n-1})$$ follows the estimate $$b^n-a^n<(b-a)nb^{n-1}$$ I can't understand why $a$ must be greater than zero.

Answer 1

The assertion "if $0\lt a\lt b$, then $b^n-a^n\lt(b-a)nb^{n-1}$" (where $n\gt1$ is a natural number), does not say that $a$ must be greater than $0$ in order for the inequality to hold, merely that the inequality holds if $a$ is greater than $0$ (and less than $b$). It would have been OK to replace the assumption $0\lt a\lt b$ with $0\le a\lt b$ or even $|a|\lt b$. The proof that is suggested here, which boils down to using the triangle inequality

$$|b^{n-1}+b^{n-2}a+\cdots+a^{n-1}|\le b^{n-1}+b^{n-2}|a|+\cdots+|a|^{n-1}$$

(followed by a strict inequality when all the $|a|$'s are replaced by $b$'s, when $n\gt1$), is unchanged.

It can be of interest to explore conditions on $a$ and $b$ for which the (strict) inequality $b^n-a^n\lt(b-a)nb^{n-1}$ does not hold, but I leave that to the OP, if it's of interest to them to do so, perhaps as a separate question.