# Find the maximum value a sum of $6$ terms

Given $6$ non-negative reals $x_1,x_2,x_3,x_4,x_5,x_6$ such that $x_1+x_2+x_3+x_4+x_5+x_6=1$ and $x_1x_3x_5+x_2x_4x_6 \geq \frac{1}{540}$.

Find the maximum value of $x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+x_4x_5x_6+x_5x_6x_1+x_6x_1x_2$.

First of all, since the required expression is not symmetric, it leads to problems. Also, I am not able to deal $540$ term. Some hints please. Thanks.

• Can all have equal denominator?? As getting $1$ wouod then be tough – Archis Welankar Dec 17 '15 at 10:27
• Sorry, what do you mean by equal denominator? – user167045 Dec 17 '15 at 10:37
• Means like $x/8,y/8..$ something like that. – Archis Welankar Dec 17 '15 at 10:43
• At least the problem has a cyclic symmetry. – Redundant Aunt Dec 17 '15 at 11:35

Hint. The expression to be maximized is $(x_1 + x_4)(x_2 + x_5)(x_3 + x_6) - (x_1 x_3 x_5 + x_2 x_4 x_6)$. Answer: $19/540$.