已知xyz∈R+求证:(1+x²)(1+y²)(1+z²)≥8xyz

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已知xyz∈R+求证:(1+x²)(1+y²)(1+z²)≥8xyz已知xyz∈R+求证:(1+x²)(1+y²)(1+z²)≥8xyz已知

已知xyz∈R+求证:(1+x²)(1+y²)(1+z²)≥8xyz
已知xyz∈R+求证:(1+x²)(1+y²)(1+z²)≥8xyz

已知xyz∈R+求证:(1+x²)(1+y²)(1+z²)≥8xyz
(1+x²)≥2x,(1+y²)≥2y,(1+z²)≥2z,
因为(1+x²)≥1为正数,所以相乘没问题
所以得证