Minimum of $\frac{1}{x+y+z}+\frac{1}{x+y+w}+\frac{1}{x+z+w}-\frac{2}{x+y+z+w}$ Let $x,y,z,w\geq 0$ and $0\leq x+y,y+z,z+x,x+w,y+w,z+w\leq 1$. What is the minimum of $$F(x,y,z,w)=\frac{1}{x+y+z}+\frac{1}{x+y+w}+\frac{1}{x+z+w}-\frac{2}{x+y+z+w}?$$
We have $F(1/2,1/2,1/2,1/2)=F(1,0,0,0)=1$, so the minimum is at most $1$. Since the constraints are on $x+y,y+z$, etc. instead of $x,y,z,w$, taking partial derivative with respect to $x,y,z,w$ doesn't help much. We cannot change one variable without affecting the others. Moreover, it is not easy to create new variables $a=x+y,b=y+z,\ldots$  and write $F$ nicely in terms of $a,b,\ldots$
 A: Did you see my last contribution in response to your rather similar previous question. You need to think a bit before asking. Here is a brief hint for isolating $w$
$$\frac 1{x+y+w}+\frac 1{x+z+w}-\frac 2 {x+y+z+w}=$$$$=\frac 1{x+y+w}-\frac 1 {x+y+z+w}+\frac 1{x+z+w}-\frac 1 {x+y+z+w}$$
If you can't make progress from there, I don't think you have understood what I put for your previous question.
A: Not a very easy problem, but the following solution is nice enough ;)
Denote $a=y+z,b=z+w,c=y+w$ then $y=\frac{a+c-b}{2},z=\frac{a+b-c}{2},w=\frac{b+c-a}{2}$. 
The constraints become
\begin{align}
x\ge 0 \\
a+b-c\ge 0 \\
b+c-a\ge 0\\
c+a-b\ge 0 \\
0\le a,b,c \le 1 \\
x+ \frac{a+b-c}{2} \le 1 \\
x+ \frac{b+c-a}{2} \le 1 \\
x+ \frac{c+a-b}{2} \le 1
\end{align}
and the function $F$ becomes
$$F = \frac{1}{a+x} + \frac{1}{b+x} + \frac{1}{c+x} - \frac{2}{x+\frac{a+b+c}{2}}.$$
WLOG, suppose that $c=\min(a,b,c)$.
We will show that $F\ge 1$. 
From the well-known inequality $\frac{1}{p}+\frac{1}{q}+\frac{1}{r} \ge \frac{9}{p+q+r} \quad \forall p,q,r > 0$ with equality iff $p=q=r$ (which follows from AM-GM inequality: $(p+q+r)\left(\frac{1}{p}+\frac{1}{q}+\frac{1}{r}\right) \ge 3\sqrt[3]{pqr}\cdot 3 \sqrt[3]{\frac{1}{pqr}}=9$), we have
$$F\ge \frac{9}{(a+x)+(b+x)+(c+x)} - \frac{4}{2x+a+b+c},$$
or $F\ge f(x)$ where $$f(x) = \frac{9}{3x+s} - \frac{4}{2x+s}$$ and $s=a+b+c$.
It is easy to show that $f(x)$ is decreasing on $[0,+\infty)$ by taking its derivative, or by just re-writing it as $$f(x) = \frac{2s}{(3x+s)(2x+s)} + \frac{3}{3x+s}.$$
Thus, since $x\le 1-\frac{a+b-c}{2}$ we have
\begin{align}
f(x) \ge f\left(1-\frac{a+b-c}{2}\right) &= \frac{9}{3\left(1-\frac{a+b-c}{2}\right) + a+b+c} - \frac{4}{2\left(1-\frac{a+b-c}{2}\right) + a+b+c} \\
&=\frac{18}{6+5c-a-b} - \frac{2}{c+1} \\
&\ge \frac{18}{6+3c} - \frac{2}{c+1} \quad (\text{since }a+b\ge 2c) \\
&= 1 + \frac{c(1-c)}{(c+1)(c+2)} \\
&\ge 1.
\end{align}
Equality occurs if and only if $a=b=c$ and $x=1-\frac{a+b-c}{2}$ and $(c=0 \text{ or } c=1)$, or equivalently, $(x,a,b,c)=(1,0,0,0)$ or $(x,a,b,c)=(1/2,1,1,1)$, i.e., $(x,y,z,w)=(1,0,0,0)$ or $(x,y,z,w)=(1/2,1/2,1/2,1/2)$.
We are done.
