已知x>y,求证:x³-y³>2x²y-2xy²
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已知x>y,求证:x³-y³>2x²y-2xy²
已知x>y,求证:x³-y³>2x²y-2xy²
已知x>y,求证:x³-y³>2x²y-2xy²
证明:
x>y
x³-y³-(2x²y-2xy²)
=(x-y)(x²+xy+y²)-2xy(x-y)
=(x-y)(x²+xy+y²-2xy)
=(x-y)(x²-xy+y²)
=(x-y)[(x-y/2)²+3y²/4]
因为:x>y
所以:x-y>0
因为:(x-y/2)²+3y²/4>0
所以:x³-y³-(2x²y-2xy²)=(x-y)[(x-y/2)²+3y²/4]>0
所以:x³-y³>2x²y-2xy²