2013 USAJMO Problems/Problem 6

Solution with Thought Process

Without loss of generality, let $x \le y \le z$ . Then $\sqrt{x + xyz} = \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$ .

Suppose x = y = z. Then $\sqrt{x + x^3} = 3\sqrt{x-1}$ , so $x + x^3 = 9x - 9$ . It is easily verified that $x^3 - 8x + 9 = 0$ has no solution in positive numbers greater than 1. Thus, $\sqrt{x + xyz} \ge \sqrt{x - 1} + \sqrt{y - 1} + \sqrt{z - 1}$ for x = y = z. We suspect if the inequality always holds.

Let x = 1. Then we have $\sqrt{1 + yz} \ge \sqrt{y-1} + \sqrt{z-1}$ , which simplifies to $1 + yz \ge y + z - 2 + 2\sqrt{(y-1)(z-1)}$ and hence $yz - y - z + 3 \ge 2\sqrt{(y-1)(z-1)}$ Let us try a few examples: if y = z = 2, we have $3 > 2$ ; if y = z, we have $y^2 - 2y + 3 \ge 2(y-1)$ , which reduces to $y^2 - 4y + 5 \ge 0$ . The discriminant (16 - 20) is negative, so in fact the inequality is strict. Now notice that yz - y - z + 3 = (y-1)(z-1) + 2. Now we see we can let $u = \sqrt{(y-1)(z-1)}$ ! Thus, $u^2 - 2u + 2 = (u-1)^2 + 1 > 0$ , and the claim holds for x = 1.

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2013 USAJMO Problems/Problem 6

Solution with Thought Process