Let, $$a \geq b \geq c > 0$$ 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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