If a &#x2265;<!-- ≥ --> b &#x2265;<!-- ≥ --> c &gt; 0 be real numbers such that <

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 ${\displaystyle \frac{a^n}{b^n+c^n}}\ge {\displaystyle \frac{a^n}{2b^n}}={\displaystyle \frac{1}{2}}{\left({\displaystyle \frac{a}{b}}\right)}^n\to \mathrm{\infty }$ as ${\displaystyle \frac{a}{b}}>1$.
Thus there exists some large $N$ such that the quotient is larger than 1, and this completes the proof by contrapositive.