设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx
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设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx
设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx
设X Y Z属于R,求证x²+y²+z²≥xy+yz+zx
∵2(x²+y²+z²)-2(xy+yz+xz)
=(x²+y²-2xy)+(y²+z²-2yz)+(x²+z²-2xz)
=(x-y)²+(y-z)²+(x-z)²
≥0
∴2(x²+y²+z²)≥2(xy+yz+xz)
∴x²+y²+z²≥xy+yz+zx