对于任意∈R,求2x^2+3xy+2y^2=1求k=x+y+xy的最小值
问题描述：
对于任意∈R,求2x^2+3xy+2y^2=1求k=x+y+xy的最小值
答
设a = x+yb = xy则k = a +b2x^2+3xy+2y^2=1 => 2a^2 - b = 1=> b = 2a^2 - 1=> k = a + 2a^2 - 1 = 2(a+1/4)^2 - 9/8 >= -9/8等号成立时 a = -1/4 b = -7/8 => a^2 - 4b >0 =>x,y能解出来 => -9/8就是k的最小值...