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Indian National Mathematical Olympiad

Algebra Difficulty 4.5 AIME Prove it India

Problem:
Let $a, b, c, x, y, z$ be positive real numbers such that $a + b + c = x + y + z$ and $a b c = x y z$ . Further, suppose that $a \leq x < y < z \leq c$ and $a < b < c$ . Prove that $a = x$ , $b = y$ and $c = z$ .

Solution

Solution:
Let
$f(t) = (t - x)(t - y)(t - z) - (t - a)(t - b)(t - c)$
Then $f(t) = k t$ for some constant $k$ . Note that $k a = f(a) = (a - x)(a - y)(a - z) \leq 0$ and hence $k \leq 0$ . Similarly, $k c = f(c) = (c - x)(c - y)(c - z) \geq 0$ and hence $k \geq 0$ . Combining the two, it follows that $k = 0$ and that $f(a) = f(c) = 0$ . These equalities imply that $a = x$ and $c = z$ , and then it also follows that $b = y$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.