If $a \geq b \geq c>0$ be real numbers such that $\forall n \in N$, there exists triangles of side lengths $a^n,b^n,c^n$.Then prove that those triangles must be Isosceles.
I am not sure how to prove this, but what i understood from the question is, Given a scalene triangle its impossible that $\forall n \in N$ the sides $a^n,b^n,c^n$ also forms scalene triangle.
First of all, i considered to eliminate Scalene Right triangles, since if $a>b>c$ forms a right scalene triangle we have $a^2+b^2=c^2$ and hence triangle inequality will not be satisfied for $a^2,b^2,c^2$ and hence they wont form a triangle at all.
Also its impossible to form a triangle with squares of sides of Isosceles triangle with sides $a,a,c$ with $c \geq \sqrt{2}a$.
But for Non right angled scalene triangles i am not sure to find $n \in N$ such that $a^n,b^n,c^n$ does not form a triangle.
Suppose the triangle is not isosceles. Then $a > b > c > 0$.
We need to show that for sufficiently large $n$, $a^n > b^n + c^n$, which violates the triangle inequality.
Note that $\dfrac {a^n}{b^n + c^n} \ge \dfrac {a^n}{2b^n} = \dfrac12\left(\dfrac ab\right)^n \to \infty$ as $\dfrac ab > 1$.
Thus there exists some large $N$ such that the quotient is larger than $1$, and this completes the proof by contrapositive.