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India — Team Selection Test, 2009

Algebra Difficulty 4.6 AIME Prove it India

Let $x$ , $y$ , $z$ and $a$ , $b$ , $c$ be positive real numbers such that $x + y + z = a + b + c$ and $xyz = abc$ . Suppose $\max\{x, y, z\} \ge \max\{a, b, c\}$ . Prove that
$ab + bc + ca \ge xy + yz + zx.$

Solution

We may assume $x \ge y \ge z$ and $a \ge b \ge c$ . We are given $x \ge a$ . Consider the cubics
$P(t) = (t - x)(t - y)(t - z), \quad Q(t) = (t - a)(t - b)(t - c).$
Since $a$ is the largest root of $Q(t) = 0$ , it follows that $Q(s) \ge 0$ for $s \ge a$ . In particular, $Q(x) \ge 0$ . Observe that
$P(t) - Q(t) = (\alpha - \beta)t,$
where $\alpha = xy + yz + zx$ and $\beta = ab + bc + ca$ . We thus get
$(\alpha - \beta)x = P(x) - Q(x) = -Q(x) \le 0.$
Since $x$ is positive, we conclude that $\alpha \le \beta$ . This gives the desired inequality.

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