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Beginners' Competition

Algebra Difficulty 4.5 AIME Prove it Austria

Prove that all real numbers x≠−1x \neq -1, y≠−1y \neq -1 with xy=1xy = 1 satisfy the following inequality:
(2+x1+x)2+(2+y1+y)2≥92 \left(\frac{2+x}{1+x}\right)^2 + \left(\frac{2+y}{1+y}\right)^2 \ge \frac{9}{2}

Solution

Since xy=1xy = 1, we may assume that x≠0x \neq 0 and y≠0y \neq 0. By substituting y=1xy = \frac{1}{x} we achieve
(2+x1+x)2+(2+y1+y)2=(2+x1+x)2+(2x+1x+1)2=5x2+8x+5x2+2x+1 \left(\frac{2+x}{1+x}\right)^2 + \left(\frac{2+y}{1+y}\right)^2 = \left(\frac{2+x}{1+x}\right)^2 + \left(\frac{2x+1}{x+1}\right)^2 = \frac{5x^2 + 8x + 5}{x^2 + 2x + 1}
and it remains to show that
5x2+8x+5x2+2x+1≥92 \frac{5x^2 + 8x + 5}{x^2 + 2x + 1} \ge \frac{9}{2}
This inequality is equivalent to the inequality
(x−1)2≥0 (x-1)^2 \ge 0
and hence everything is proved.

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