Sequence of triangles with exponentiated sides

Question

Let $a,b,c\in\Bbb R^+$. For any given $n\in\Bbb N$ there exists a triangle with sides $a^n,b^n,c^n$. Show that at least two of $a,b,c$ are equal.

I know that you can create inequalities for $a^n,b^n,c^n$, but what's next?

Answer 1

If all of $a,b,c$ are different, without loss of generality let $a<b<c$. Define $x=\frac ab$ and $y=\frac cb$ so that we have $x<1<y$.

Since $x<1$, for all $n\in\Bbb N$ we have $x^n<1$ and hence $x^n+1^n=x^n+1<2$. However, since $y>1$, $y^n$ grows without bound as $n$ increases, and in particular there must be an $n$ where $y^n>2$. At this exponent we must have $$x^n+1^n<y^n$$ Multiplying by $b^n$: $$a^n+b^n<c^n$$ So $a^n,b^n,c^n$ are not the sides of a triangle and the triple $(a,b,c)$ does not satisfy the question's condition. Therefore, at least two of $a,b,c$ have to be equal.

Answer 2

Assume WLOG that $c > b > a$, then we have by the Triangle Inequality that: $a^n + b^n > c^n$. Dividing by $c^n$ we have that $\left(\frac ac \right)^n + \left(\frac bc \right)^n > 1$. But we have that $\frac ac, \frac bc < 1$, meaning that the LHS can get arbitrarily close to $0$. But this violates the Triangle Inequality, so our assumption is wrong. Meaning that at least one of the sides must be equal.