Question

[pham trung thanh]

Áp dụng BĐT Cô-si:

  \(a^2+b^2\ge2ab\)

\(b^2+c^2\ge2bc\)

\(c^2+a^2\ge2ca\)

\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)

\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Answer 1

[Trần Quốc Anh]

Other way:\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca\ge3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca\ge0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca\ge0\)

\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)(đúng)

Dấu "=" xảy ra khi a=b=c