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...
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Áp dụng BĐT Cô-si:
\(a^2+b^2\ge2ab\)
\(b^2+c^2\ge2bc\)
\(c^2+a^2\ge2ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)\ge2\left(ab+bc+ca\right)\)
\(\Rightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
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