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.