Bài 1:
Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=-\left(a^2+b^2+c^2\right)\)
Ta thấy \(\left\{{}\begin{matrix}a^2\ge0\\b^2\ge0\\c^2\ge0\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge0\Rightarrow-\left(a^2+b^2+c^2\right)\le0\)
\(\Rightarrow2\left(ab+bc+ca\right)\le0\)
\(\Leftrightarrow ab+bc+ca\le0\left(đpcm\right)\)
Vậy...
Với \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ac+bc+ac\right)\)
Vì \(a^2\ge0;b^2\ge0;c^2\ge0\)(với mọi a,b,c\(\in\)R)
\(\Rightarrow\)\(a^2+b^2+c^2\ge0\) (đẳng thức xảy ra khi a=b=c=0)
\(\Rightarrow-2\left(ab+bc+ac\right)\ge0\)
\(\Rightarrow ab+bc+ac\le0\)(đpcm)
