Minimum of $\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}$ Let $0\leq x,y,z\leq 1$. What is the minimum of $$F(x,y,z)=\frac{1}{x+y}+\frac{1}{x+z}-\frac{1}{x+y+z}?$$
We have $F(1,1,1)=2/3$, $F(1,1,0)=F(1,0,1)=F(1,0,0)=1$, and $F(0,1,1)=3/2$.
Is the minimum $2/3$?

This can be done in an elementary way.
Consider the last two terms, which contain $z$ - they sum to $$\frac y{(x+z)(x+y+z)}$$
Increasing $z$ clearly decreases this term, so for a minimum you want $z$ as large as possible - changing $z$ doesn't change the first term at all. You can do a similar analysis for $y$. So with the constraints you have $y=z=1$
$$F(x,1,1)=\frac 2{x+1}-\frac 1{x+2}=\frac 1{x+1}+\frac 1{(x+1)(x+2)}$$
This is the sum of two positive terms which decrease with increasing $x$ (in the range you have) so the minimum is when $x=1$.
