已知x^2+y^2=1,x^2+z^2=2,y^2+z^2=2,求xy+xz+yz的最小值
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已知x^2+y^2=1,x^2+z^2=2,y^2+z^2=2,求xy+xz+yz的最小值
已知x^2+y^2=1,x^2+z^2=2,y^2+z^2=2,求xy+xz+yz的最小值
已知x^2+y^2=1,x^2+z^2=2,y^2+z^2=2,求xy+xz+yz的最小值
x^2+y^2=1,x^2+z^2=2,y^2+z^2=2
三式相加
2(x²+y²+z²)=5
x²+y²+z²=5/2
所以 x²=y²=1/2,z²=3/2
x=±√2/2,y=±√2/2,z=±√6/2
因为求 xy+xz+yz的最小值,所以 x,y,z只能是有正有负
总共八种情形,去掉x,y,z全正,全负的情形,
xy+xz+yz=xy+(x+y)z
(1) x=√2/2,y=√2/2,z=-√6/2 xy+xz+yz=1/2- √3
(2) x=√2/2,y=-√2/2,z=-√6/2 xy+xz+yz=-1/2
(3) x=√2/2,y=-√2/2,z=√6/2 xy+xz+yz=-1/2
(4) x=-√2/2,y=√2/2,z=-√6/2 xy+xz+yz=-1/2
(5) x=-√2/2,y=√2/2,z=√6/2 xy+xz+yz=-1/2
(6) x=-√2/2,y=-√2/2,z=√6/2 xy+xz+yz=1/2- √3
所以 ,最小值为1/2- √3
x²+z²=2
y²+z²=2
两式相减,得:x=y
又x²+y²=1,则:x²=y²=1/2,从而z²=3/2
xy+xz+yz=x²+2xz=(1/2)-2×(√2/2)×(√6/2)=(1/2)-(√3)=[1-2√3]/2
则:xy+yz+zx的最小值是[1-2√3]/2
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