Question

CMR: 3(a^3+b^3+c^3)>= (a+b+c)(a^2+b^2+c^2);

Answer 1

\(3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)\left(a^2+b^2+c^2\right)\)

\(\Leftrightarrow3\left(a^3+b^3+c^3\right)\ge a^3+ab^2+ac^2+ba^2+b^3+bc^2+ca^2+cb^2+c^3\)

\(\Leftrightarrow2a^3+2b^3+2c^3\ge ab^2+ac^2+ba^2+bc^2+ca^2+cb^2\)

Lại có:\(a^2-ab+b^2\ge ab\)(dễ dàng cm đc khi chuyển vế)

\(\Rightarrow\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)

\(\Rightarrow a^3+b^3\ge ab\left(a+b\right)\)

TT\(\Rightarrow b^3+c^3\ge bc\left(b+c\right);a^3+c^3\ge ac\left(a+c\right)\)

Cộng vế theo vế =>đpcm