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Ta có: $xyz-\frac{16}{x+y+z}=0\Leftrightarrow xyz=\frac{16}{x+y+z} \Leftrightarrow x+y+z=\frac{16}{xyz}$ (1)$\Rightarrow Q=(x+y)(x+z)=x^{2}+xy+xz+yz=x(x+y+z)+yz=x.\frac{16}{xyz}+yz=\frac{16}{yz}.yz\geq 2.\sqrt{\frac{16}{yz}.yz}=2.4=8$.$\Rightarrow Q_{min}=8$Dấu = xảy ra khi$yz=\frac{16}{yz}\Leftrightarrow (yz)^{2}=16\Leftrightarrow yz=4($Vì$x,y,z>0)$

Ta có: $xyz-\frac{16}{x+y+z}=0\Leftrightarrow xyz=\frac{16}{x+y+z} \Leftrightarrow x+y+z=\frac{16}{xyz}$ (1)$\Rightarrow Q=(x+y)(x+z)=x^{2}+xy+xz+yz=x(x+y+z)+yz=x.\frac{16}{xyz}+yz=\frac{16}{yz}+yz\geq 2.\sqrt{\frac{16}{yz}.yz}=2.4=8$.$\Rightarrow Q_{min}=8$Dấu = xảy ra khi$yz=\frac{16}{yz}\Leftrightarrow (yz)^{2}=16\Leftrightarrow yz=4($Vì$x,y,z>0)$

Ta có: $xyz-\frac{16}{x+y+z}=0\Leftrightarrow xyz=\frac{16}{x+y+z} \Leftrightarrow x+y+z=\frac{16}{xyz}$ (1)$\Rightarrow Q=(x+y)(x+z)=x^{2}+xy+xz+yz=x(x+y+z)+yz=x.\frac{16}{xyz}+yz=\frac{16}{yz}.yz\geq 2.\sqrt{\frac{16}{yz}.yz}=2.4=8$.$\Rightarrow Q_{min}=8$

Ta có: $xyz-\frac{16}{x+y+z}=0\Leftrightarrow xyz=\frac{16}{x+y+z} \Leftrightarrow x+y+z=\frac{16}{xyz}$ (1)$\Rightarrow Q=(x+y)(x+z)=x^{2}+xy+xz+yz=x(x+y+z)+yz=x.\frac{16}{xyz}+yz=\frac{16}{yz}.yz\geq 2.\sqrt{\frac{16}{yz}.yz}=2.4=8$.$\Rightarrow Q_{min}=8$Dấu = xảy ra khi $yz=\frac{16}{yz}\Leftrightarrow (yz)^{2}=16\Leftrightarrow yz=4$ $($Vì $x,y,z>0)$