Ta có: \(a^2+b^2+c^2=ab+ac+bc\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+ac+bc\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)
=> đpcm.
\(a^2+b^2+c^2=ab+ac+bc\Rightarrow2a^2+2b^2+2c^2=2ab+2ac+2bc\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Rightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
\(\Rightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Mà \(\left(a-b\right)^2\ge0;\left(a-c\right)^2\ge0;\left(b-c\right)^2\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(b-c\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\Rightarrow a=b=c\)
Vậy a=b=c