\(a^2+b^2+c^2+3=2\left(a+b+c\right)\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\)
Lời giải:
Áp dụng BĐT Cauchy:
\(\left\{\begin{matrix} a^2+1\geq 2|a|\geq 2a\\ b^2+1\geq 2|b|\geq 2b\\ c^2+1\geq 2|c|\geq 2c\end{matrix}\right.\Rightarrow a^2+b^2+c^2+3\geq 2(a+b+c)\)
Dấu bằng xảy ra khi:
\(\left\{\begin{matrix} a,b,c\geq 0\\ a^2=1\\ b^2=1\\ c^2=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a=1\\ b=1\\ c=1\end{matrix}\right.\)
