$(a+b+c)^{3}\geq a^{3}+b^{3}+c ^{3} + 24abc$

Đáp án:

Có `:` `( a + b + c )^3 = a^3 + b^3 + c^3 + 3 ( a + b ) ( b + a ) ( c + a )`

Mà `:` `a^3 + b^3 + c^3 + 3 ( a + b ) ( b + a ) ( c + a ) ≥ a^3 + b^3 + c^3 + 24abc`

Ta lại có `:`

`( a + b ) ( b + c ) ( a + c ) - 8abc`

`= a^2c + abc + ac^2 + a^2b + ab^2 + abc + b^2c + bc^2 - 8abc`

`= c ( a - b )^2 + a ( b - c )^2 + b ( c - a )^2 ≥ 0`

`->` `( a + b ) ( b + c ) ( a + c ) ≥ 8abc`

`->` `3 ( a + b ) ( b + c ) ( a + c ) ≥ 24abc`

`->` `( a + b + c )^3 ≥ a^3 + b^3 + c^3 + 24abc` `(` đpcm `)`