`color{#1be01b}{-T}color{#1be01b}{i}color{#1be01b}{t}color{#1be01b}{u}color{#1be01b}{u}color{#1be01b}{u-}color{#1be01c}`

Với `a; b; c` không âm ta có `(a+b+c)/(3)>=\root{3}{abc}`

Đặt `\root{3}{a}=x; \root{3}{b}=y; \root{3}{c}=z (x; y; z>=0)`

`-> x^3+y^3+z^3>=3xyz`

`-> [(x+y)^3+z^3]-3xy(x+y)-3xyz>=0`

`-> (x+y+z)[(x+y)^2-(x+y)z+z^2]-3xy(x+y+z)>=0`

`-> (x+y+z)(x^2+y^2+z^2+2zy-xz-yz-3zy)>=0`

`-> (x+y+z)(x^2+y^2+z^2-xy-yz-zx)>=0`

`-> (x+y+z)(2z^2+2y^2+2z^2-2zy-2yz-2zx)>=0`

`-> (x+=y+z)[(x-y)^2+(y-2)^2+(z-x)^2]>=0 AA x; y; z`

Dấu `=` xảy ra khi `x=y=z` hay `a=b=c`

Vậy `AA a; b; c>=0` thì `a+b+c)/(3)>=\root{3}{abc}`