\(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
=> \(a^2+b^2+c^2=ab+ac+bc\)
=> \(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
=> \(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
=> \(\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\)
=> a = b = c
Vậy .....