Competition problem — Algebra — Maths Olympiad Prep

Library / /347 of 520

Algebra Difficulty 5.8 AIME, harder Prove it

[ Algebraic problems on the triangle inequality ]

For any natural number $n$ , from the numbers $a^{\mathrm{n}}, b^{\mathrm{n}}$ and $c^{\mathrm{n}}$ a triangle can be formed. Prove that among the numbers $a, b$ and $c$ there are two equal.

Solution

We can assume that $a \geq b \geq c$ . Let's prove that $a=b$ . Indeed, if $b<a$ , then $b \leq \lambda a$ and $c \leq \lambda a$ , where $\lambda<1$ . Therefore, $b^{\mathrm{n}}+c^{\mathrm{n}} \leq 2 \lambda^{n} a^{\mathrm{n}}$ . For sufficiently large $n$ , we have $2 \lambda^{n}<1$ and we obtain a contradiction with the triangle inequality.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.