\(a^2+b^2+c^2=ab+bc+ac\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)(1)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}}\forall a,b,c\)\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\left(\forall a,b,c\right)\)(2)
Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\a-c=0\end{cases}}\Rightarrow a=b=c\)
Vậy \(a=b=c\)
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)
\(\Leftrightarrow a=b=c\)
