\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Ta có: \(a^2+b^2+c^2+3=2.\left(a+b+c\right)\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}}\text{vì }\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
\(\Leftrightarrow a=b=c=1\)
Vậy a=b=c=1
Ta có: a2+b2+c2+ 3=2(a+b+c)
<=>\(\hept{\begin{cases}a=1\\b=1\\c=1\end{cases}}\)
