X大于0,Y大于0 X+Y+XY=2 求X+Y的最小值

1个回答

  • 解法一:

    x>0,y>0,则依二元均值不等式得:

    2=x+y+xy

    ≤(x+y)+(x+y)²/4

    →(x+y)²+4(x+y)-8≥0

    →(x+y+2+2√3)(x+y-2√3)≥0.

    而x+y+2+2√3>0,

    ∴x+y+2-2√3≥0,

    ∴(x+y)|min=-2+2√3.

    解法二:

    设x+y=t(x>0,y>0,即t>0)

    代入条件式则

    t+(t-x)x=2

    →x²-tx+2-t=0

    判别式不小于0,即

    (-t)²-4(2-t)≥0

    →t²+4t-8≥0

    →(t+2+2√3)(t+2-2√3)≥0.

    因t>0→t+2+2√3>0,

    ∴t+2-2√3≥0.

    ∴(x+y)|min=-2+2√3.