An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$ 
Let $a, b, c, d \ge 0$ such that $a+b, a+c, a+d, b+c, b+d, c+d \le 1$. Show that
$$2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd).$$

I am trying to maximise
$$f(a,b,c,d)=(a+b+c+d) /(1-(1-a)(1-b)(1-c)(1-d) +abcd) $$ over the set of nonnegative real numbers $a, b, c, d$ subject to $a+b, a+c, a+d, b+c, b+d, c+d \le 1$.
When $a=b=c=d=0$, set $f(0,0,0,0)=1$. Some massage with Wolfram Alpha gives $2$ as a local maximum, which, if it were the global maximum, would then be equivalent to the above inequality.
It seems one standard way to solve inequalities of this class is the pqrs, uvwt method.
 A: The form of the constraints on the variables suggest a transform based around their differences from each other and from $1/2$. Let $a = (1-x+y)/2$, $b = (1 - x - y)/2$, $c = (1-z+w)/2$, $d = (1-z-w)/2$. The positivity constraints are $x + |y| \le 1$ and $z + |w| \le 1$. The sum constraints are $x,z\in[0,1]$ and $|y|+|w|\le x+z$.
With these substitutions, the inequality can be reduced to
$$
(x+z)(1-xz) + zy^2 + xw^2 \ge 0,
$$
which is clearly true whenever $x,z\in [0,1]$.
This substitution also works with your function method. The expression for the function becomes
$$
\frac{4(2-x-z)}{4 - (x+z)(xz+1)+xw^2 + zy^2}.
$$
This function's value can only be increased by decreasing $|y|$ and $|w|$. Additionally, decreasing $|y|$ or $|w|$ can never cause the constraints to become unsatisfied. Thus its maximum value is will be the same as its maximum value assuming $y = w = 0$. That is, $a = b$ and $c = d$. The constraints then require $0\le a,c\le 1/2$. It's not hard to show the maximum value of $f(a,c,a,c)$ subject to these constraints is $2$, occurring at $a = c = 1/2$.
A: Remark: Here is a proof which is similar to my answer
in An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$
Let $x = a + b, \, y = c + d$.
We have $x, y \in [0, 1]$.
We have
\begin{align*}
 \mathrm{RHS} - \mathrm{LHS} &= (x + y) + 2(ab y + cdx) - 2(ab + cd + xy)\\
 &= (x + y) - 2xy - 2(1 - y)ab - 2(1 - x)cd\\
 &\ge (x + y) - 2xy - 2(1 - y)\cdot \frac{x^2}{4} - 2(1 - x)\cdot \frac{y^2}{4} \tag{1}\\
 &= (x + y) - 2xy - x^2/2 - y^2/2 + x^2y/2 + xy^2/2 \\
 &= (x + y) - xy - (x + y)^2/2 + xy(x + y)/2 \\
 &= (x + y) - (x + y)^2/2 -\frac{1}{2}(2 - x - y)xy\\
 &\ge (x + y) - (x + y)^2/2 -\frac{1}{2}(2 - x - y)\cdot \frac{(x + y)^2}{4}\\
 &= \frac18(x + y)(2 - x - y)(4 - x - y)\\
 &\ge 0
\end{align*}
where we have used $ab \le (a + b)^2/4 = x^2/4$ and $cd \le (c + d)^2/4 = y^2/4$ in (1).
We are done.
