Prove that $c^n>a^n+b^n$ for $n>2$

Question

It is given that $a,b,c$ are the sides of a triangle.Also $c^2=a^2+b^2$.Prove that $c^n>a^n+b^n$ for $n\in\mathbb N$ such that $n>2$.

I dont have any idea on how to solve this.please help.

Answer 1

You have $$ \left(\frac{a}{c}\right)^2+\left(\frac{b}{c}\right)^2=1$$

Use this to conclude that $\left(\frac{a}{c}\right),\left(\frac{b}{c}\right) \lt1$ and that $$\left(\frac{a}{c}\right)^n \lt \left(\frac{a}{c}\right)^2 \\ \left(\frac{b}{c}\right)^n \lt \left(\frac{b}{c}\right)^2$$

You need then the $c$ positive constrain to deduce your inequality.

P.S. Deleting and reposting the same question is not nice. I already typed this answer to the deleted question, to find out that I cannot post it!

Answer 2

$c^n=c^{n-2}c^2=c^{n-2}(a^2+b^2)=a^2 c^{n-2}+b^2 c^{n-2}>a^2a^{n-2}+b^2b^{n-2}=a^n+b^n$

Using $a,b<c$ as $a,b,c$ are sides of a right angled triangle.