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Algebra Difficulty 4.6 AIME Prove it Croatia

Let xx, yy and zz be positive real numbers such that x+y+z=18xyzx + y + z = 18xyz. Prove the inequality
xx2+2yz+1+yy2+2xz+1+zz2+2xy+1≥1. \frac{x}{\sqrt{x^2 + 2yz + 1}} + \frac{y}{\sqrt{y^2 + 2xz + 1}} + \frac{z}{\sqrt{z^2 + 2xy + 1}} \ge 1.

Solution

The A-H inequality together with the condition of the problem gives
xy+yz+zx=xyz(1x+1y+1z)≥xyz⋅9x+y+z=9xyz18xyz=12. xy + yz + zx = xyz \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) \ge xyz \cdot \frac{9}{x + y + z} = \frac{9xyz}{18xyz} = \frac{1}{2}.
Using 1≤2xy+2yz+2zx1 \le 2xy + 2yz + 2zx we get
x2+2yz+1≤x2+2xy+2zx+4yz=(x+2y)(x+2z)≤(x+y+z)2 x^2 + 2yz + 1 \le x^2 + 2xy + 2zx + 4yz = (x + 2y)(x + 2z) \le (x + y + z)^2
where the last inequality is a consequence of the A-G inequality. Hence xx2+2yz+1≥xx+y+z\frac{x}{\sqrt{x^2 + 2yz + 1}} \ge \frac{x}{x + y + z} and
analogously yy2+2zx+1≥yx+y+z\frac{y}{\sqrt{y^2 + 2zx + 1}} \ge \frac{y}{x + y + z}, zz2+2xy+1≥zx+y+z\frac{z}{\sqrt{z^2 + 2xy + 1}} \ge \frac{z}{x + y + z}.
Adding these three inequalities finishes the proof.

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