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.

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.
