∴x,y>0
lg(x+y)=lgx+lgy=lgxy
∴x+y=xy
When 0 < x ≤ 1, x+y=xy≤y, which is not true.
∴x> 1
∴y=x/(x- 1)= 1+ 1/(x- 1)
Obviously, y=f(x) is a decreasing function on the interval (1,+infinity).
x+y = x+ 1+ 1/(x- 1)= x- 1+ 1/(x- 1)+2≥2 √( x- 1)× 1/(x- 1)+2 = 4
∴x+y≥4