因为x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx) =(x+y+z)[(x-y)^2+(y-z)62+(z-x)^2]/2>=0,所以 x^3+y^3+z^3>=3xyz,对于 a + b + 1/(ab) ；且 a >0,b > 0 设 x = a^(1/3),y = b^(1/3),z = (1/ab)^(1/3) 则 利用...