Ta có \(a+b+c=0\)
\(=>a=-b-c\)
Ta có \(ab+bc+ac\le0\)
\(=>\left(-b-c\right)b+bc+\left(-b-c\right)c\le0\)
\(=>-b^2-bc+bc-bc-c^2\le0\)
\(=>-b^2-bc-c^2\le0\)
\(=>-\left(b^2+bc+c^2\right)\le0\)(ĐPCM)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(a^2+b^2+c^2\ge0\)
\(a^2+b^2+c^2=-\left(2ab+2bc+2ac\right)\)
\(\Rightarrow2ab+2bc+2ca\le0\Leftrightarrow ab+bc+ac\le0\)