实数x,y,z,若x2+y2=1,y2+z2=2,z2+x2=2,则xy+yz+zx的最小值是怎求_数学解答

实数x,y,z,若x2+y2=1,y2+z2=2,z2+x2=2,则xy+yz+zx的最小值是怎求

人气：170 ℃　时间：2019-12-13 23:17:46

解答

x2+y2=1,y2+z2=2,z2+x2=2三式相加,可得x²+y²+z²=(1+2+2)/2=5x²+y²+z²+(xy+yz+zx)=(1/2)[(x+y)²+(y+z)²+(z+x)²]>=0xy+yz+zx>=-(x²+y²+z²)=-5/2xy+yz+zx...