If $ABC$ is a right angled triangle at $C$ then prove that $a^n+b^n<c^n$ for all $n>2$

Question

If $ABC$ is a right angled triangle at $C$ then prove that $a^n+b^n<c^n$ for all $n>2$

This is an olympiad book problem. I know that $a+b>c$ and $a^2+b^2=c^2$, $a^n+b^n≠c^n$ for all $n>2$, and $(a+b)^n>c^n$ but I cant find a way to put these together and prove the above, may be I am not hitting up the right approach .......Need help.

Answer 1

If $n>2$, the unit ball of the norm $\|(x,y)\|_2=(|x|^2+|y|^2)^{1/2}$ is contained in the unit ball of $\|(x,y)\|_n=(|x|^n+|y|^n)^{1/n}$, and there are only four common points: $(\pm 1,0)$ and $(0,\pm 1)$. Therefore, if $(a/c)^2+(b/c)^2=1$, then the point $(a/c,b/c)$ lies in the interior of the unit ball of $\|\cdot\|_n$, because $a,b,c$ are all non-zero.

Answer 2

Well, we know $a^2+b^2 = c^2$ and we know $a\lt c$ and $b\lt c$ so we can write this:

$$a^n+b^n = a^2a^{n-2}+b^2b^{n-2} \leq a^2(\max\{a, b\})^{n-2}+b^2(\max\{a, b\})^{n-2} \lt (a^2+b^2)c^{n-2} = c^n$$