设x,y,z属于R求证x的平方加y的平方加z的平方大于等于xy+yz+zx 若x的平方+y的平方+z的平方-xy-yz-zx=0,求证:...
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\u6240\u4ee5x�0�5+y�0�5+z �0�5\u2265xy\uff0byz\uff0bzx
x²+y²+z²=xy+yz+zx
x²+y²+z²-xy-yz-xz=0
\u4e24\u8fb9\u4e582
2x²+2y²+2z²-2xy-2yz-2xz=0
(x²-2xy+y²)+(y²-2yz+z²)+(z²-2xz+x²)=0
(x-y)²+(y-z)²+(z-x)²=0
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\u6240\u4ee5x-y=0,y-z=0,z-x=0
x=y,y=z\uff0cz=x
\u6240\u4ee5x=y=z
所以
x^2+y^2≥2xy
x^2+z^2≥2xz
y^2+z^2≥2yz
相加得
2（x^2+y^2+z^2)≥2(xy+yz+zx)
所以x^2+y^2+z^2≥xy+yz+zx
∵(x-y)²≥0，(y-z)²≥0，(z-x)²≥0
∴x²+y²≥2xy，y²+z²≥2yz，z²+x²≥2zx
∴(x²+y²)+(y²+z²)+(z²+x²)≥2xy+2yz+2zx
∴2(x²+y²+z²)≥2(xy+yz+zx)
∴(x²+y²+z²)≥(xy+yz+zx)
x²＋y²≥2xy；y²＋z²≥2yz；z²＋x²≥2xz，三个式子相加，得：
2x²＋2y²＋2z²≥2xy＋2yz＋2zx
即：
x²＋y²＋z²≥xy＋yz＋zx
(x-y)^2+(y-z)^2+(z-x)^2≥0
x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2≥0
2(x^2+y^2+z^2)-2(xy+yz+zx)≥0
2(x^2+y^2+z^2)≥2(xy+yz+zx)
x^2+y^2+z^2≥xy+yz+zx
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