Question

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

Answer 1

由已知条件可得x^2=y^2=1/2,z^2=3/2.因为xy+yz+zx= xy +(x+y)z,

当x,y同号时,xy=x^2=1/2,xy+yz+zx ≥ 1/2 - 2/√2 * √(3/2) = 1/2 - √3.

当x,y异号时,xy=-1/2,x+y=0,xy+yz+zx =-1/2 >1/2 - √3.

所以当x=y=1/√2,z=-√(3/2)时 xy+yz+zx的最小值是1/2 - √3.