Maximum Product Value 1
Let $ x_1, x_2, \dots , x_6 $ be non-negative real numbers such that $ x_1 +x_2 +x_3 +x_4 +x_5 +x_6 =1 $, and $ x_1 x_3 x_5 +x_2 x_4 x_6 \ge \frac{1}{540} $. Find the maximum value of
\[x_1 x_2 x_3 + x_2 x_3 x_4 +x_3 x_4 x_5 +x_4 x_5 x_6 +x_5 x_6 x_1 +x_6 x_1 x_2.\]
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Solution
Let $ a = x_1 x_3 x_5 + x_2 x_4 x_6 $ and $ b = x_1 x_2 x_3 + x_2 x_3 x_4 + x_3 x_4 x_5 + x_4 x_5 x_6 + x_5 x_6 x_1 + x_6 x_1 x_2 $. By AM-GM,
\[a + b = (x_1 + x_4)(x_2 + x_5)(x_3 + x_6) \le \left[ \frac{(x_1 + x_4) + (x_2 + x_5) + (x_3 + x_6)}{3} \right]^3 = \frac{1}{27}.\]Hence,
\[b \le \frac{1}{27} - \frac{1}{540} = \frac{19}{540}.\]Equality occurs if and only if
\[x_1 + x_4 = x_2 + x_5 = x_3 + x_6.\]We also want $ a = \frac{1}{540} $ and $ b = \frac{19}{540} $. For example, we can take $ x_1 = x_3 = \frac{3}{10}, $ $ x_5 = \frac{1}{60}, $ $ x_2 = \frac{1}{3} - x_5 = \frac{19}{60}, $ $ x_4 = \frac{1}{3} - x_1 = \frac{1}{30}, $ and $ x_6 = \frac{1}{3} - x_3 = \frac{1}{30} $.
Thus, the maximum value of $ b $ is $ \boxed{\frac{19}{540}} $.
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