Do x,y,z$\geq$0&x+y+z=1$\Rightarrow$0$\leq$x,y,z$\leq$1(1)Ta có:2P=(2x+4y+6z)(6x+3y+2z)$\leq$$(\frac{8(x+y+z)-y}{2})^{2}$=$(\frac{8-y}{2})^{2}$
Từ (1)$\Rightarrow$$(\frac{8-y}{2})^{2}$$\leq$$\frac{8^{2}}{4}$=16$\Rightarrow$P$\leq$8
Dấu''='' xra$\Leftrightarrow$x=z=$\frac{1}{2}$&y=0