10.1. Different positive numbers $x, y, z$ satisfy the equations

$$
x y z=1 \quad \text{and} \quad x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}
$$

Find the median of them. Justify your answer.
# Solution:

Method 1. Consider the polynomial $P(t)=(t-x)(t-y)(t-z)$. The numbers $x, y, z$ are its roots. Expanding the brackets or using Vieta's theorem for cubic polynomials, we get $P(t)=t^{3}+a t^{2}+b t+c$, where $a=-(x+y+z)$, $b=xy+xz+yz$, and $c=-xyz$. The condition of the problem shows that $a=-b$ and $c=-1$. Then we have $P(t)=t^{3}-b t^{2}+b t-1=(t-1)(t^{2}-b t+1)$. This means that one of the roots of the polynomial $P(t)$ is 1, and the other two are the roots of the quadratic trinomial $t^{2}-b t+1$. By Vieta's theorem, their product is 1. Since both are positive numbers, one of them is greater than 1, and the other is less than 1.

Method 2. Express $z$ from the first equation and substitute it into the second: $z=\frac{1}{xy}$, and then

\[
\begin{gathered}
x+y+\frac{1}{xy}=\frac{1}{x}+\frac{1}{y}+xy \\
x^{2}y+xy^{2}+1=y+x+x^{2}y^{2} \\
x^{2}(y-y^{2})+x(y^{2}-1)+1-y=0 \\
(1-y)(x^{2}y-xy-x+1)=0 \\
(1-y)(x-1)(xy-1)=0
\end{gathered}
\]

Thus, either $x=1$, or $y=1$, or $xy=1 \Leftrightarrow z=1$. In any of these cases, the product of the other two numbers is 1, as follows from the first condition. Since the numbers are distinct and positive, one of them is less than 1, the second is equal to 1, and the third is greater than 1.

Answer: 1.

Note: Reference to Vieta's theorem is accepted without proof.

| in progress | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| The problem is reduced to finding the roots of a cubic equation | 2 points |
| The problem is reduced to an equation with two variables | 1 point |
| An example of numbers satisfying the conditions of the problem is provided | 0 points |
| Incorrect solution or its absence | 0 points |