Let $$ a\geq b \geq c > 0 $$ be real numbers such that for all $$n $$ element of natural number , there exists triangles of side length $$ a^n , b^n , c^n. $$ Prove that the triangles are isosceles .
I tried taking derivatives and proving that there will exist some $n$ at which inequality would stop holding. But it is not working out.
Also I tried substituting triangle inequalities for $n=1$ ; but it is of no use I guess.
I think binomial maybe of some use. I tried completing the binomial of $a^n + b^n$. But I am not able to use it too.
If the triangle with sides $a,b,c$ were not isosceles, then $a>b>c>0$. Then, for $n$ sufficiently big, $a^n>b^n+c^n$, against the assumption that $a^n,b^n$ and $c^n$ are the lengths of the sides of a triangle. It means that at least one equality of $a\geq b\geq c$ must hold (actually, we can even say that $a=b$).