Ta có: a2 + b2 + c2 - ab - bc - ca = 0
=> aa + bb + cc - ab - bc - ca = 0
=> aa + ab - bb + bc - cc -+ca = 0
=> a - b - c = 0
=> a = b = c (đpcm)
a2+b2+c2-ab-bc-ca=2a2+2b2+2c2-2ab-2bc-2ca=a2-2ab+b2+b2-2bc+c2+c2-2cb+b2=(a-b)2+(b-c)2+(c-a)2=0
=>\(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}=>\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}=>a=b=c\)
ta có \(a^2+b^2+c^2-ab+bc-ba=0\)
suy ra\(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\)\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Do \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)
