Consider a minimizer $(a,b,x,y)$ (which exists by continuity plus compactness). (1) If $x\le y$, then $b+x\le b+y\le1$ and $a+x\le a+y\le2$. In this case the minimum value is 3, and is attained for instance for $x=y=b=1/2$ and $a=3/2$. (2) In the remaining cases $y\le x$. The constraints imply $a,x\le2$ and $b,y\le1$. The minimizer satisfies $a=2-x$. (2a) If $1\le x-y$, then $2-x\le 1-y$. The minimizer satisfies $b=2-x$, and hence $a=b=2-x$. The considered value becomes $\frac12+\frac{1}{2-x+y}+\frac12+\frac{1}{2-x+y}$. The assumption $1\le x-y$ yields $2-x+y\le1$. Hence the value is at most 3, and is attained for instance for $a=b=y=1/2$ and $x=3/2$. (2b) If $x-y\le1$, then $1-y\le2-x$. The minimizer satisfies $b=1-y$. The considered value becomes $\frac12+\frac{1}{2-x+y}+\frac{1}{1-y+x}+1$. We set $z:=x-y$, and minimize $f(z)=\frac{1}{2-z}+\frac{1}{1+z}$ subject to $0\le z\le1$. This auxiliary minimum is $4/3$, as $f(z)\ge4/3$ is equivalent to $(2z-1)^2\ge0$. Hence the value is at most $17/6$, and is attained for instance for $b=y=1/2$ and $a=x=1$. The summarizing answer: the minimum is $17/6$.

Denote $(a+x,a+y,b+x,b+y)=(p,q,r,s)$. From the condition we have: $p,q,r\le 2, \; s\le 1 \quad (*)$ and $q+r=p+s\le 3 \quad (**)$. Hence, $(*)$ gives: $\frac{1}{p}+\frac{1}{s}\ge \frac{3}{2} \quad (1)$ and am-hm with $(**)$ implies: $\frac{1}{q}+\frac{1}{r}\ge \frac{4}{q+r}\ge \frac{4}{3} \quad (2)$ Therefore, summing $(1)$ and $(2)$, we find that minimum value is $\frac{17}{6}$ ,with equality case, for example: $(a,b,x,y)=\left(\frac{1}{2},0,\frac{3}{2},1\right)$

Of course, not $\dfrac {1}{2} + \dfrac {1}{2} + \dfrac {1}{2} + 1 = \dfrac {5}{2}$. Say $a+x = s \leq 2$, $a+y = t \leq 2$, $b+x = u \leq 2$, $b+y = v \leq 1$. Then $s+v = t+u \leq 3$, so $(a+y) + (b+x) \leq 3$. Thus $\dfrac {1}{a+y} + \dfrac {1}{b+x} \geq \dfrac {4}{(a+y)+(b+x)} \geq \dfrac {4}{3}$, with equality for $a+y = b+x = 3/2$. But $\dfrac {1}{a+x} + \dfrac {1}{b+y} \geq \dfrac {1}{2} + 1$, with equality for $a+x = 2$, $b+y = 1$. Since this system has solution, for example $a=x=1$, $b=y=1/2$, the minimal value of that sum is $\dfrac {4}{3} + \dfrac {1}{2} + 1 = \dfrac {17}{6} > \dfrac {5}{2}$.

MathUniverse wrote: with equality case, for example: $(a,b,x,y)=\left(\frac{1}{2},0,\frac{3}{2},1\right)$ Minus one point!

danepale wrote: Determine the lowest possible value of the expression$\frac{1}{a+x} + \frac{1}{a+y} + \frac{1}{b+x} + \frac{1}{b+y} $, where $a,b,x,$ and $y$ are positive real numbers satisfying the inequalities $\frac{1}{a+x} \ge \frac{1}{2} ,\frac{1}{a+y} \ge \frac{1}{2},\frac{1}{b+x} \ge \frac{1}{2} ,\frac{1}{b+y} \ge 1. $ $a+b+ x+y\le 3$, $\frac{1}{a+x} + \frac{1}{a+y} + \frac{1}{b+x} + \frac{1}{b+y} \ge \frac{1}{2} + \frac{4}{a+y+ b+x} + 1$ $ \ge \frac{1}{2} + \frac{4}{3} + 1= \frac{17}{6},$ with equality for$a+x=2,b+y=1,a+y=b+x=\frac{3}{2} .$