jmc

The equations are equivalent to $$\begin{array}{rl}4x^2-4y+1& =0,\\ 4y^2-4x+1& =0.\end{array}$$ Summing these equations gives $$4x^2-4y+1+4y^2-4x+1=0,$$ or $$(4x^2-4x+1)+(4y^2-4y+1)=0.$$ Factoring the squares of binomials gives $$(2x-1{)}^2+(2y-1{)}^2=0.$$ Since squares are always non-negative, it follows that $$2x-1=2y-1=0,$$ so $x=y=\frac{1}{2}$. The desired answer is $\frac{1}{\frac{1}{8}+\frac{1}{8}}=\boxed{4}$.