Ta có:
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2-2\left(a+b+c\right)+3=0\)
\(\Leftrightarrow a^2+b^2+c^2-2a-2b-2c+3=0\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)
\(\Leftrightarrow a=b=c=1\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\\ \Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\\ \Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
ta luôn có: \(\left(a-1\right)^2\ge0;\left(b-1\right)^2\ge0;\left(c-1\right)^2\ge0\\ \Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
đẳng thức xảy ra khi và chỉ khi :
\(\left\{{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\Rightarrow a=b=c=1\left(đpcm\right)\)
