已知x,y,z∈R+,x²+y²+z²≤（根号3）xyz,求证：x+y+z≤xyz

Question

已知x,y,z∈R+,x²+y²+z²≤（根号3）xyz,求证：x+y+z≤xyz

Answer 1

首先,由均值不等式,有x²+y²+z² ≥ 3(x²y²z²)^(1/3).
代入条件得3√3·(xyz)³ = (√3·xyz)³ ≥ (x²+y²+z²)³ ≥ 27(xyz)².
进而由(xyz)² > 0可得xyz ≥ 3√3.
其次,由Cauchy不等式,有3(x²+y²+z²) ≥ (x+y+z)².
代入条件得3√3·xyz ≥ 3(x²+y²+z²) ≥ (x+y+z)².
进而由xyz ≥ 3√3可得(xyz)² ≥ (x+y+z)².
又xyz > 0,x+y+z > 0,故xyz ≥ x+y+z.
注:Cauchy不等式那一步等价于x²+y²+z² ≥ xy+yz+zx,
也可以由(x-y)²+(y-z)²+(z-x)² ≥ 0展开得到.