Giả sử: \(x^2+4y^2+3z^2+14>2x+12y+6x\)
\(\Leftrightarrow\left(x^2-2x+1\right)+\left(4y^2-12y+9\right)+3\left(z^2-2x+1\right)+1\)> 0
\(\Leftrightarrow\left(x-1\right)^2+\left(2y-3\right)^2+3\left(z-1\right)^2+1>0\) (luôn đúng).
Suy ra: \(x^2+4y^2+3z^2+14>2x+12y+6x\).