AM-GM
$\frac {1}{\sqrt {(1 - x)(1 - y)}} \ge \frac {2}{2 - x - y}$
$\frac {2}{9}(\frac {1}{\sqrt {(1 - a)(1 - b)}} + \frac {1}{\sqrt {(1 - a)(1 - c)}} + \frac {1}{\sqrt {(1 - a)(1 - d)}} + \frac {1}{\sqrt {(1 - b)(1 - c)}} + \frac {1}{\sqrt {(1 - b)(1 - d)}} + \frac {1}{\sqrt {(1 - c)(1 - d)}}) \ge$
$\frac {2}{9}( \frac {2}{2 - a - b} + \frac {2}{2 - a - c} + \frac {2}{2 - a - d} + \frac {2}{2 - b - c} + \frac {2}{2 - b - d} + \frac {2}{2 - c - d})$
$= \frac {2}{9}( \frac {2}{a + b} + \frac {2}{a + c} + \frac {2}{a + d} + \frac {2}{b + c} + \frac {2}{b + d} + \frac {2}{c + d}).$
จาก โคชี จะได้ว่า $\frac {2}{a + b} + \frac {2}{a + c} + \frac {2}{a + d} \ge \frac {18}{3a + b + c + d} = \frac {18}{2} (\frac {1}{a + 1})$
และ $\frac {2}{a + b} + \frac {2}{b + c} + \frac {2}{b + d} \ge \frac {18}{3b + a + c + d} = \frac {18}{2} (\frac {1}{b + 1})$
$\frac {2}{c + b} + \frac {2}{a + c} + \frac {2}{c + d} \ge \frac {18}{3c + b + a + d} = \frac {18}{2} (\frac {1}{c + 1})$
และ $\frac {2}{a + d} + \frac {2}{d + c} + \frac {2}{b + d} \ge \frac {18}{3b + b + c + a} = \frac {18}{2} (\frac {1}{d + 1})$
นำอสมการที่ได้มาบวกกันจะได้ว่าอสมการที่โจทย์ต้องการเป็นจริง
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