If $x_1+x_2+\cdots x_8=1$, then show that $x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_5x_6+x_6x_7+x_7x_8+x_8x_5+x_1x_5+x_2x_6+x_3x_7+x_4x_8\le 1/4$. If $x_1+x_2+\cdots x_8=1$ and $x_1,\ldots x_8$ are all non-negative numbers, then show that $x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_5x_6+x_6x_7+x_7x_8+x_8x_5+x_1x_5+x_2x_6+x_3x_7+x_4x_8\le 1/4$.
I'd like to interpret this prolem as follows: Think of $x_1,\cdots x_8$ as 8 vertices of a cube and 12 products as 12 edges of a cube. Each product is the product of two vertices on that edge. Does this interpretation do any help? Or is there any other way to think about this problem?
 A: You have done well to identify the cubic idea. Now look at how this appears in terms of the two tetrahedra in the cube, by which I mean you get the expression $$(x_1+x_3+x_6+x_8)(x_2+x_4+x_5+x_7)-x_1x_7-x_2x_8-x_3x_5-x_4x_6$$$$=y(1-y)-x_1x_7-x_2x_8-x_3x_5-x_4x_6$$ where $y=x_1+x_3+x_6+x_8\ge 0$
Suppose we have chosen wlog $x_1\ge x_3, x_6, x_8\ge 0$ to be an optimising set.  Then first we note that we can set $x_7=0$ because $x_1x_7 \ge x_2x_7$ so that $x_1x_7+x_2x_8\ge x_1\cdot 0 +x_2(x_7+x_8)$.
Then we note that with $x_7=0$ we have $x_2, x_4, x_5\ge 0 (=x_7)$ so that $x_2x_8+x_3x_5+x_4x_6\ge (x_1+x_8+x_3+x_6)\cdot x_7=y\cdot x_7=0$ And this means that an optimum choice is $x_1=y$ (which minimises the amount we deduct) and $x_3=x_6=x_8=0$.
Then (best case) the trailing products are all zero and it is easy to optimise $y=0.5$. Note that because the sketch above mass choices in cases of equality, there are many solutions with $y=0.5$ and the trailing products all zero - I will leave it to you to see how they relate to this cubic expression of the inequality which you have picked out (the trailing products are opposite vertices of the cube).
