设u=x-1>0,则x=u+1,u+y=1,1/(x-1)+1/(2y)=[1/u+1/(2y)](u+y)=1+u/(2y)+y/u+1/2≥3/2+√2，当u/(2y)=y/u，即y=√2-1，u=2-√2时取等号，所以所求的最小值是3/2+√2.
