Saudi Arabia booklet, 2024

Algebra Difficulty 4.9 AIME Prove it Saudi Arabia

Let $x$ , $y$ , $z$ be real numbers such that $x \ge y \ge z \ge 0$ and $2x + y + 2z = 5$ . Prove that
$5 \le x^2 + z^2 + xy + yz + zx \le \frac{25}{4}.$
When does the equality case hold?

Solution

Let $P = x^2 + z^2 + xy + yz + zx$ then we have
$25 = (2x + y + 2z)^2 = 4x^2 + y^2 + 4z^2 + 4xy + 4yz + 4zx = 4P + y^2 + 4zx.$
Note that $y^2 + 4zx \ge 0$ so $4P \le 25$ which implies that $P \le \frac{25}{4}$ . The equality case is $y = z = 0$ and $x = \frac{5}{2}$ .

Continue, note that $(y - x)(y - z) \le 0$ so $y^2 + zx \le xy + yz$ . Thus
$P \ge x^2 + z^2 + y^2 + 2zx = (x + z)^2 + y^2.$
Using Cauchy-Schwarz inequality, one can get
$(2^2 + 1^2)((x+z)^2 + y^2) \ge (2x + 2z + y)^2 = 25.$
Thus $(x+z)^2 + y^2 \ge \frac{25}{5} = 5$ which implies that $P \ge 5$ .
The equality case is $x = y = z = 1$ . $\square$

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