Find the smallest possible value of

$$
x y+y z+z x+\frac{1}{x}+\frac{2}{y}+\frac{5}{z},
$$

for positive real numbers $x, y$ and $z$.
Answer: the smallest possible value is $3 \sqrt[3]{36}$.
Using the arithmetic-geometric mean inequality, we find

$$
\begin{aligned}
x y+\frac{1}{3 x}+\frac{1}{2 y} & \geq 3 \sqrt[3]{x y \frac{1}{3 x} \frac{1}{2 y}}=3 \sqrt[3]{\frac{1}{6}} \\
y z+\frac{3}{2 y}+\frac{3}{z} & \geq 3 \sqrt[3]{y z \frac{3}{2 y} \frac{3}{z}}=3 \sqrt[3]{\frac{9}{2}} \\
x z+\frac{2}{3 x}+\frac{2}{z} & \geq 3 \sqrt[3]{x z \frac{2}{3 x} \frac{2}{z}}=3 \sqrt[3]{\frac{4}{3}}
\end{aligned}
$$

When we add these three inequalities, we get

$$
x y+y z+z x+\frac{1}{x}+\frac{2}{y}+\frac{5}{z} \geq 3\left(\sqrt[3]{\frac{1}{6}}+\sqrt[3]{\frac{9}{2}}+\sqrt[3]{\frac{4}{3}}\right)=3\left(\frac{1}{6}+\frac{1}{2}+\frac{1}{3}\right) \sqrt[3]{36}=3 \sqrt[3]{36}
$$

For each of the three inequalities, equality holds when the three terms are equal. We arrive at an equality case for all three when $(x, y, z)=\left(\frac{1}{3} \sqrt[3]{6}, \frac{1}{2} \sqrt[3]{6}, \sqrt[3]{6}\right)$, which is also easy to verify. Then $x y+y z+z x+\frac{1}{x}+\frac{2}{y}+\frac{5}{z}$ is indeed equal to

$$
\frac{1}{6} \sqrt[3]{36}+\frac{1}{2} \sqrt[3]{36}+\frac{1}{3} \sqrt[3]{36}+3 \frac{1}{\sqrt[3]{6}}+4 \frac{1}{\sqrt[3]{6}}+5 \frac{1}{\sqrt[3]{6}}=\sqrt[3]{36}+12 \frac{1}{\sqrt[3]{6}}=\sqrt[3]{36}+2 \sqrt[3]{36}=3 \sqrt[3]{36}
$$

Remark. After deciding to split the inequality into three inequalities using the arithmetic-geometric mean (one for $x y$, one for $y z$, and one for $x z$), there are different ways to find the coefficients for $\frac{1}{x}, \frac{1}{y}$, and $\frac{1}{z}$. First, you can be guided by examples you have found in your search for the equality case. For example, the expression equals 10 for $(x, y, z)=\left(\frac{2}{3}, 1,2\right)$, and with $(x, y, z)=\left(\frac{3}{5}, \frac{9}{10}, \frac{9}{5}\right)$, you get $\frac{324}{100}+\frac{12}{18} \cdot 10=3.24+6.6666 \ldots=9.9066 \ldots$.

A more structured approach is to introduce variables $a, b, c, d, e, f$ for the inequalities on $x y+a \frac{1}{x}+b \frac{1}{y}, y z+c \frac{1}{y}+d \frac{1}{z}$, and $z x+e \frac{1}{z}+f \frac{1}{x}$. Since these three terms must add up to the desired expressions, we get the conditions

$$
\begin{aligned}
f+a & =1, \\
b+c & =2, \\
d+e & =5 .
\end{aligned}
$$

Furthermore, we want equality to hold for certain $x, y$, and $z$. So the system

$$
\begin{aligned}
& x y=a \frac{1}{x}=b \frac{1}{y} \\
& y z=c \frac{1}{y}=d \frac{1}{z} \\
& z x=e \frac{1}{z}=f \frac{1}{x}
\end{aligned}
$$

must have a solution for some $x, y$, and $z$. From the first equation, it follows that $\frac{x}{y}=\frac{a}{b}$. But using the last two equations, we can also write this as

$$
\frac{x}{y}=\frac{z x}{y z}=\frac{f / x}{c / y}=\frac{f y}{c x} .
$$

This means that $\frac{f}{c}=\frac{x^{2}}{y^{2}}=\frac{a^{2}}{b^{2}}$, or $a^{2} / f=b^{2} / c$. Similarly, $c^{2} / b=d^{2} / e$ and $e^{2} / d=f^{2} / a$ hold. Together with $f+a=1, b+c=2$, and $d+e=5$, we now have six equations in six variables. Although it is not easy to solve this system, it greatly simplifies the search because it is now easy to check whether $a, b, c, d, e, f$ satisfy the conditions. Other equations that can be derived from this system include $ace = bdf$ and $ad = be = cf$.