An algebra problem by Rui-Xian Siew

$\begin{aligned} Q & = \frac {(\color{#3D99F6}{x+y})(y+z)(z+x)}{xyz} \\ & = \frac {(\color{#3D99F6}{1-z})(1-x)(1-y)}{xyz} & \small \color{#3D99F6}{\text{Note that }x+y+z =1} \\ & = \frac {1-(\color{#3D99F6}{x+y+z})+xy+yz+zx-xyz}{xyz} \\ & = \frac {1-\color{#3D99F6}{1}+xy+yz+zx-xyz}{xyz} \\ & = \color{#3D99F6}{\frac 1x + \frac 1y + \frac 1z} - 1 & \small \color{#3D99F6}{\text{By AM-HM inequality (see Note)}} \\ & \ge \color{#3D99F6}{\frac 9{x+y+z}} - 1 = \boxed{8} \end{aligned}$

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Note: AM-HM inequality states that

$\begin{aligned} \frac {x+y+z}3 \ge \frac 3{\frac 1x + \frac 1y + \frac 1z} \\ \implies \frac 9{x+y+z} \le \frac 1x + \frac 1y + \frac 1z \end{aligned}$

Call the expression A. By AM-GM we get $x+y\geq 2\sqrt{xy}$ , $y+z\geq 2\sqrt{yz}$ , $x+z\geq 2\sqrt{xz}$ . So $A\geq\frac{8xyz}{xyz}=8$ The equality holds at $x=y=z=\frac{1}{3}$