a2 + b2 + c2 = ab + ac + bc
<=> 2a2 + 2b2 + 2c2 = 2ab + 2ac + 2bc
<=> a2 - 2ab + b2 + b2 - 2bc + c2 + a2 - 2ac + c2 = 0
<=> (a - b)2 + (b - c)2 + (a - c)2 = 0
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)
<=> a = b = c
\(a^2+b^2+c^2=ab+ac+bc\\ a^2+b^2+c^2-ab-ac-bc=0\\ 2a^2+2b^2+2c^2-2ab-2ac-2bc=0\\ a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\\ \left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\left(đpcm\right)\)
Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\) \(a^2+a^2+b^2+b^2+c^2+c^2-2ab-2bc=2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+b^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\)
\(\Rightarrow a=b=c\) (đpcm)
