Question

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

Answer 1

根据完全平方公式因为x2＋y2大于等于2XY 所以XY大于等于2分之1又因为都是加法 成递增 所以xy＋yz＋zx的最小值就是3者相加等于2分之5

Answer 2

∵x�0�5+y�0�5 ≥2xy y�0�5+z�0�5≥2yz x�0�5+z�0�5≥2xz∴2（x�0�5+y�0�5+z�0�5） ≥ 2（xy+yz+xz)即：x�0�5+y�0�5+z�0�5 ≥xy+yz+xz ∵已知的三式相加，得；2x�0�5 +2y�0�5+2z�0�5 = 5 即：x�0�5+y�0�5+z�0�5 =5/2∴xy+yz+xz ≤ 5/2

Answer 3

x2+y2=1,
y2+z2=2,
z2+x2=2
三式相加,可得
x�0�5+y�0�5+z�0�5=(1+2+2)/2=5

x�0�5+y�0�5+z�0�5+(xy+yz+zx)
=(1/2)[(x+y)�0�5+(y+z)�0�5+(z+x)�0�5]>=0
xy+yz+zx>=-(x�0�5+y�0�5+z�0�5)=-5/2

xy+yz+zx>=-5/2
xy+yz+zx的最小值是-5/2

Answer 4

x2+y2=1→x2+y2+2xy=1+2xy→(x+y)^2=1+2xy
y2+z2=2→y2+z2+2yz=2+2yz→ (y+z)^2=2+2yz
z2+x2=2→z2+x2+2xz=2+2xz→ (x+z)^2=2+2xz

(x+y)^2+(y+z)^2+(x+z)^2=1+2xy+2+2yz+2+2xz
(x+y)^2+(y+z)^2+(x+z)^2=5+2(xy+yz+zx)
(xy+yz+zx)=【(x+y)^2+(y+z)^2+(x+z)^2-5】÷2

因为（x+y)^2+(y+z)^2+(x+z)^2大于或等于0，
所以，(xy+yz+zx)的值大于或等于-5/2