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If $a\ge b\ge c>0$ be real numbers such that $\mathrm{\forall }n\in N$, there exists triangles of side lengths ${a}^{n},{b}^{n},{c}^{n}$. Then prove that those triangles must be Isosceles.
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Kiana Cantu
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 $\frac{{a}^{n}}{{b}^{n}+{c}^{n}}\ge \frac{{a}^{n}}{2{b}^{n}}=\frac{1}{2}{\left(\frac{a}{b}\right)}^{n}\to \mathrm{\infty }$ as $\frac{a}{b}>1$.
Thus there exists some large $N$ such that the quotient is larger than 1, and this completes the proof by contrapositive.