Q:

Maximum of $(ab+cd)(ac+bd)(ad+bc)$

Let $a,b,c,d\ge 0$ satisfy $a+b+c+d=4$. Find the maximum value of $(ab+cd)(ac+bd)(ad+bc)$.
When all of the variables are $1$, the value is $8$. Using the AM-GM inequality gives $$(ab+cd)(ac+bd)(ad+bc)\leq\left(\frac{ab+cd+ac+bd+ad+bc}{3}\right)^3.$$
Can we upper bound the right-hand side in terms of $a+b+c+d$?

A:

We have 
$$ab+ac+ad+bc+bd+cd=\frac12\bigl((a+b+c+d)^2-(a^2+b^2+c^2+d^2)\bigr) $$
and for $a+b+c+d=4$ we have $a^2+b^2+c^2+d^2\ge4$ (why?) so that 
$$ab+ac+ad+bc+bd+cd\le\frac{16-4}{2}= 6 $$
and you indeed obtain $\le 8$ for the original inequality.

