Exercise 5. Let $x, y, z$ be strictly positive real numbers such that

$$
x+\frac{y}{z}=y+\frac{z}{x}=z+\frac{x}{y}=2
$$

Determine all possible values that the number $x+y+z$ can take.
Solution to Exercise 5 In such a problem, it is necessary to examine each equation separately but also to relate them. In practice, this involves looking at the equation obtained when performing the sum or product of two or more equations. Another idea is to apply known inequalities on one side or the other of the equation. Indeed, often the equalities present in the problems are only possible when the variables satisfy the equality case of a well-known inequality.

Let's start with a separate examination. We consider the first equation $x+\frac{y}{z}=2$. According to the inequality of means, we have

$$
2=x+\frac{y}{z} \geqslant 2 \sqrt{x \cdot \frac{y}{z}}
$$

which means $\sqrt{\frac{x y}{z}} \leqslant 1$. Similarly, the second equation $y+\frac{z}{x}=2$ gives $\sqrt{\frac{y z}{x}} \leqslant 1$. By multiplying these two inequalities, we obtain

$$
1 \geqslant \sqrt{\frac{x y}{z}} \cdot \sqrt{\frac{y z}{x}}=y
$$

Thus $y \leqslant 1$, and similarly we find that $x \leqslant 1$ and $z \leqslant 1$. We now seek to apply these estimates or to prove inverse estimates (for example, $y \geqslant 1$, which would imply that $y=1$).

Note that the relation $x+\frac{y}{z}=2$ can be rewritten as $x z+y=2 z$. Similarly, we get $y x+z=2 x$ and $y z+x=2 y$. By summing these three relations, we find that $x z+y x+y z+x+y+z=2(x+y+z)$, or equivalently $x y+y z+z x=x+y+z$. But then

$$
x+y+z=x y+y z+z x \leqslant x \cdot 1+y \cdot 1+z \cdot 1=x+y+z
$$

Our inequality is actually an equality. And since $x, y$, and $z$ are non-zero, each of the inequalities used in the line above is actually an equality. Therefore, we have $x=y=z=1$ and $x+y+z=3$. This value is indeed achievable since the triplet $(1,1,1)$ satisfies the system given in the problem, which completes our problem.

## Comments from the graders:

The exercise was quite well done. The methods of solving are extremely varied. Some students made the mistake of considering the problem symmetric, whereas it is only cyclic.