Let $a\ge b \ge c\gt 0$ be real numbers such that for all $n \in \mathbb N$ there exist triangles of side $a^n, b^n, c^n$. Prove that the triangles are isosceles.
I tried proving it by writing $c^n + b^n \gt a^n$ and when I assumed some values for $a, b \text{ and } c$ I realized that it would be true for all $n$ only when $c=b$. But I don't know how to generally prove this.
On
If $a > b \ge c$ then $\frac{a}{b} > 1$ so there is some $n$ such that $\frac{a^n}{b^n} > 2$. For this value of $n$
$a^n > 2b^n \ge b^n + c^n$
so there is no triangle with sides $a^n, b^n, c^n$.
On the other hand if $a=b\ge c$ then
$a \le b+c$
$\Rightarrow a^n \le (b+c)^n \quad \forall n \in N$
$\Rightarrow a^n \le b^n + c^n \quad \forall n \in N$
so there is a triangle with sides $a^n, b^n, c^n$ for all $n \in N$.
If $a>b\geq c$ then $a^n$ will grow faster than $b^n+c^n$ and eventually the triagle inequality will be violated because $a^n>b^n+c^n$. Equivalently, $b^n+c^n$ grows slower than $a^n$. To check this compute $\lim_{n\to\infty} \frac{b^n+c^n}{a^n}$.