\(3=\frac{1}{\sqrt{3}}.\sqrt{3x}+\frac{1}{2}\sqrt{4y}+\frac{1}{\sqrt{6}}.\sqrt{6z}\)
\(\Rightarrow9\le\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\right)\left(3x+4y+6z\right)\)
\(\Leftrightarrow9\le\frac{3}{5}.\left(3x+4y+6z\right)\)
\(\Rightarrow P\ge15\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(\frac{16}{9};1;\frac{4}{9}\right)\)