Question

[Linh Tu]

Chứng minh rằng: \(a^3+b^3+c^3\ge3abc\)

Answer 1

[nguyễn thị nguyệt]

\(giải:\)

\(a^3\)\(+b^3\)\(+c^3\)\(\ge3abc\)

\(\Rightarrow a^3\)\(+b^3\)\(+c^3\)\(-3abc\ge0\)

\(\Rightarrow a^3\)\(+b^3\)\(+c^3\)\(-3abc+3a^2b+3ab^2-3a^2b-3ab^2\ge0\)

\(\Rightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-\left(3abc+3a^2b+3ab^2\right)\ge0\)

\(\Rightarrow\left(a+b\right)^3+c^3-3ab\left(c+a+b\right)\ge0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\ge0\)

\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2-3ab\right]\ge0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\ge0\)

\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\ge0\)

\(\Rightarrow\frac{1}{2}\left(a+b+c\right)\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)\ge0\)

\(\Rightarrow\frac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]\ge0\)(luôn đúng \(\forall\)a,b,c\(\ge0\))

hay \(a^3+b^3+c^3\ge3abc\left(đpcm\right)\)