sqing wrote: Let $x,y,z$ is positive. Solve: $\begin{cases}{x\left( {6 - y} \right) = 9}\\ {y\left( {6 - z} \right) = 9}\\ {z\left( {6 - x} \right) = 9}\end{cases}$ $y=\frac{6x-9}{x}$ and $z=\frac{9}{6-x}$. Hence, $\frac{6x-9}{x}\left(6-\frac{9}{6-x}\right)=9$ or $(x-3)^2=0$. Where we saw inequalities?

sqing wrote: Let $x,y,z$ is positive. Solve: $\begin{cases}{x\left( {6 - y} \right) = 9}\\ {y\left( {6 - z} \right) = 9}\\ {z\left( {6 - x} \right) = 9}\end{cases}$ Since $6>x,y,z$ $\implies xyz(6-x)(6-y)(6-z)=9^3$ $x(6-x)y(6-y)z(6-z)=9^3$ AM-GM $\prod_{cyc}\left(\frac{x+6-x}2\right)^2\geq x(6-x)y(6-y)z(6-z)=9^3$ The equality holds when $x=6-x,y=6-y,z=6-z \iff x=3,y=3, z=3$ Done!

Mathskidd wrote: sqing wrote: Let $x,y,z$ is positive. Solve: $\begin{cases}{x\left( {6 - y} \right) = 9}\\ {y\left( {6 - z} \right) = 9}\\ {z\left( {6 - x} \right) = 9}\end{cases}$ Since $6>x,y,z$ $\implies xyz(6-x)(6-y)(6-z)=9^3$ $x(6-x)y(6-y)z(6-z)=9^3$ AM-GM $\prod_{cyc}\left(\frac{x+6-x}2\right)^2\geq x(6-x)y(6-y)z(6-z)=9^3$ The equality holds when $x=6-x,y=6-y,z=6-z \iff x=3,y=3, z=3$ Done! Very nice. GMO 2017