SELECTION EXAMINATION

First solution We write the system in the form $$xy+yz+zx=x+y+z(1)$$ $$xyz=1.(2)$$ Subtracting the two equations by parts we find $$\begin{array}{rl}& xyz-(xy+yz+zx)=1-(x+y+z)\\ \iff & xyz-xy-yz-zx+x+y+z-1=0\\ \iff & xy(z-1)-x(z-1)-y(z-1)+(z-1)=0\\ \iff & (z-1)(xy-x-y+1)=0\\ \iff & (z-1)(x-1)(y-1)=0\\ \iff & x=1\text{ or }y=1\text{ or }z=1.\end{array}$$ For $x=1$, from (1) and (2) we have $yz=1$, which have the solutions $$(y,z)=\left(a,\frac{1}{a}\right),a>0,$$ and so, the solutions of the system are $$(x,y,z)=\left(1,a,\frac{1}{a}\right),a>0.$$ Similarly, considering $y=1$ or $z=1$ we find the solutions $$(x,y,z)=\left(a,1,\frac{1}{a}\right)\text{ or }(x,y,z)=\left(a,\frac{1}{a},1\right),a>0.$$ Since for each $a>0$ we have $$x+y+z=1+a+\frac{1}{a}\ge 1+2=3.$$ Equality holds for $a=1$, it follows that between the solutions of the system, the $(x,y,z)=(1,1,1)$ is that having the least possible sum $x+y+z$.

Second solution Let $(x,y,z)$ is the solution of the system with the least possible sum. Then, from the inequality of arithmetic – geometric mean we have $$\frac{x+y+z}{3}\ge \sqrt[3]{xyz}\Rightarrow x+y+z\ge 3,$$ while equality holds for $x=y=z$. Hence the least possible value of the sum $x+y+z$, between the solution of the given system is 3 and it happens for $x=y=z$. For $x=y=z$, from the equation $xyz=1$, $x,y,z>0$, it follows that $x=y=z=1$, which satisfies also the equation $x+y+z=xy+yz+zx$.