# Inequality real analysis

I'm trying to prove the following inequality: $$|a^n - b^n | \leq n M^{n-1}|a-b|$$ with $$M=\max\{|a|,|b|\}$$.

My first attempt was using mathematical induction, but I got stuck. I also tried using the triangle inequality but I can't reach the desired result.

• Hint: $a^n-b^n$ is divisible by $a-b$. – Math1000 Dec 6 '19 at 22:02
• The right hand term should probably say $nM^{n-1}|a-b|$, right? – Qi Zhu Dec 6 '19 at 22:03
$$|a^n-b^n|=|a-b|\left|\sum_{k=0}^{n-1}a^kb^{n-k-1}\right|\leq |a-b|M^{n-1}n$$
Using the mean value theorem for the function $$f(x) = x^n$$ we get that there exists some $$\theta \in \langle a,b\rangle$$ such that $$|a^n - b^n| = |f(a) - f(b)| = |f'(\theta)||a-b| = n|\theta|^{n-1}|a-b| \le nM^{n-1}|a-b|$$ since $$|\theta| \le \max\{|a|, |b|\} = M$$.
You can use the triangle inequality like this: $$|a^n-b^n|\leq |a^n-a^{n-1}b|+|a^{n-1}b-a^{n-2}b^{2}|+\cdots+|ab^{n-1}-b^n|.$$ Each term on the right can be written as $$|a-b||a|^j|b|^k$$ where $$j+k=n-1$$, and hence each term on the right is bounded above by $$M^{n-1}|a-b|$$. Now add.