\(a^2+4b^2+3c^2+14\ge2a+12b+6c\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(4b^2-12b+9\right)+3\left(c^2-2c+1\right)+1\ge0\)
BĐT \(\Leftrightarrow\left(a^2-2a+1\right)+\left(4b^2-12b+9\right)+3\left(c^2-2c+1\right)\)
\(\Leftrightarrow\left(a-1\right)^2+\left(2b-3\right)^2+3\left(c-1\right)^2\ge0\)
Dấu "=" xảy ra khi và chỉ khi : \(\left\{{}\begin{matrix}a-1=0\\2b-3=0\\c-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=\frac{3}{2}\\c=1\end{matrix}\right.\)
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