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Austrian Mathematical Olympiad

Algebra Difficulty 5.0 AIME, harder Prove it Austria

Let $x$ , $y$ and $z$ be positive real numbers with $x + y + z = 3$ . Prove that at least one of the three numbers
$x(x + y - z), \quad y(y + z - x) \quad \text{or} \quad z(z + x - y)$
is less or equal $1$ .

Solution

Since the three expressions are cyclic, we may w. l. o. g. assume that $x \ge y, z$ . Consequently we have $x \ge \frac{x+y+z}{3} = 1$ . We now show that $a := y(y+z-x) = y(3-2x)$ satisfies $a \le 1$ .

* Case a): For $\frac{3}{2} \le x < 3$ clearly $a \le 0 < 1$ .

* Case b): For $1 \le x < \frac{3}{2}$ the factor $3-2x$ is positive. Therefore $a \le x(3-2x)$ . Hence it suffices to prove $x(3-2x) \le 1$ , which is equivalent to $2x^2 - 3x + 1 \ge 0$ , i.e. $(2x-1)(x-1) \ge 0$ .

This completes the proof.

(Walther Janous) $\square$

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