解：∵x>0，y>0∴x+y≥2√xy>0∴xy=x+y+3≥2√xy+3∴xy-2√xy-3≥0(√xy+1)(√xy-3)≥0①√xy+1≥0，√xy-3≥0解得：√xy≥3②√xy+1≤0，√xy-3≤0解得：√xy≤-1(舍去)综上所述，√xy≥3∴x+4y≥2√(x·4y)=4√xy≥4×3=12∴x+4y的最小值是12。
