 Let, $a \geq b \geq 0$ be real numbers such that for all natural number n, there exist triangles of side lengths $a^n,b^n,c^n$  Prove that the triangles are isosceles.
If a, b, c are sides of a triangle, triangular inequality assures that difference of two sides is lesser than the third side. Since $a \ge b \ge c > 0$, hence using triangular inequality we have a – b < c. Infact for all n, $a^n – b^n < c^n$
We have $a^n – b^n = (a-b)(a^{n-1} + a^{n-2}b + … + b^{n-1} ) < c^n$ Replacing every a by b in the left hand side, we make the expression to the left even smaller. i.e. $(a-b)(b^{n-1} + b^{n-2}b + … + b^{n-1} ) \le (a-b)(a^{n-1} + a^{n-2}b + … + b^{n-1} ) < c^n$ Hence $(a-b) \times n \times b^{n-1} < c^n$
Now notice $(a-b) < \frac {c^n}{n\times b^{n-1}} = \frac{c}{n} \times \frac {c^{n-1}}{b^{n-1}} = \frac {c}{n} (\frac{c}{b})^{n-1}$ Clearly $\frac{c}{b} \le 1$ by given hypothesis. Hence $a-b \le \frac{c}{n}$ for all n. But letting n go to infinity, we see that a and b can be made arbitrarily close to each other. This implies a=b. Hence each triangle in the sequence is isosceles

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