MEMO 2014 T1

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 $$\frac{1}{a+x} \ge \frac{1}{2}$$ $$\frac{1}{a+y} \ge \frac{1}{2}$$ $$\frac{1}{b+x} \ge \frac{1}{2}$$ $$\frac{1}{b+y} \ge 1.$$