If $a^n, b^n$ and $c^n$ form a triangle for all $n \in \Bbb N$, prove that the triangles are isosceles.

Question

If $a^n, b^n$ and $c^n$ form a triangle for all $n \in \Bbb N$, prove that the triangles are isosceles given that $a\geq b\geq c> 0$.

I started on the lines of $a< b+c$. But no idea how to proceed further. Tried using cosine law, but no help.

Answer 1

Hint Look at the possible values of $$\lim_{n \to \infty} \frac{b^n+c^n}{a^n}$$

Answer 2

let's assume, wolog $a\ge b \ge c$.

To be a triangle we need $c^n > a^n - b^n =(a-b)(a^n + a^{n-1}b + .... + ab^{n-1} + b^n)$

If $a = b$ this is always true but if $a \ne b$. Then:

we need $c^n > a^n - b^n =$

$(a-b)(a^n + a^{n-1}b + .... + ab^{n-1} + b^n) > $

$(a-b)(a^n)$ or in other words we need

$a-b > \frac {c^n}{a^n} = (\frac ca)^n$.

but $0< \frac ca < 1$ so $\lim_{n\to \infty } (\frac ca)^n = 0$.

Which means there is an $N$ so that $(\frac ca)^N < a-b$.

So it is not possible for all $n$ for $a-b > (\frac ca)^n$.

So $a \ne b$ is impossible and $a = b$.