TITLE: Bound on $c-b$ for $a^n+b^n=c^n$
QUESTION [5 upvotes]: Let $a\leq b\leq c$ be positive real numbers and $n$ positive integer with $a^n+b^n=c^n$. Prove that $c-b\leq(\sqrt[n]{2}-1)a$.
The desired inequality can be written as $c-b+a\leq \sqrt[n]{2}a$. Raising to the power of $n$, this is $(c-b+a)^n\leq 2a^n$. If it were true that $(c-b+a)^n\leq c^n-b^n+a^n$ we would be done, since the latter is just $2a^n$.

REPLY [4 votes]: Fix $a$. By convexity of $f(x) = x^n$ on $\mathbb{R}_+$, an increase in $b$ leads to a lesser increase in $c$. Hence, the difference $c-b$ is maximized when $a=b$. The inequality follows.