u=(x-y+1)²+√(2x-y) +3
因为(x-y+1)²≥0,√(2x-y)≥0；所以u=(x-y+1)²+√(2x-y) +3≥3,
即u=(x-y+1)²+√(2x-y) +3的最小值是3.
由方程组：(x-y+1)²=0与√(2x-y)=0得到x=1,y=2.