3. Find all triples of real numbers $x, y$, and $z$ for which

$$
x\left(y^{2}+2 z^{2}\right)=y\left(z^{2}+2 x^{2}\right)=z\left(x^{2}+2 y^{2}\right) .
$$
3. If, for example, $x=0$, we obtain the system $0=y z^{2}=2 y^{2} z$, so one of the values $y, z$ is zero and the other can be arbitrary. Similarly, we solve the cases $y=0$ and $z=0$. This gives us three groups of solutions $(t, 0,0),(0, t, 0)$ and $(0,0, t)$, where $t$ is any real number. All other solutions already satisfy the condition $x y z \neq 0$, whose validity we will assume for the rest of the solution.

By rearranging the equation $x\left(y^{2}+2 z^{2}\right)=y\left(z^{2}+2 x^{2}\right)$, we get $(2 x-y)\left(z^{2}-x y\right)=0$. We thus distinguish which of the two factors is zero.

(i) $2 x-y=0$. Substituting $y=2 x$ into the original system leaves us with the single equation

$$
2 x\left(2 x^{2}+z^{2}\right)=9 x^{2} z,
$$

from which, after dividing by the number $x \neq 0$, we get

$$
4 x^{2}+2 z^{2}-9 x z=0 \text { or } (x-2 z)(4 x-z)=0 \text {. }
$$

Case (i) thus corresponds to the solution groups $(2 t, 4 t, t)$ and $(t, 2 t, 4 t)$, where $t$ is any real number.

(ii) $z^{2}-x y=0$. Partial substitution $z^{2}=x y$ gives us the single equation

$$
x y(2 x+y)=z\left(x^{2}+2 y^{2}\right) \text {, }
$$

which is (since $x^{2}+2 y^{2}>0$) equivalent to the equation

$$
z=\frac{x y(2 x+y)}{x^{2}+2 y^{2}}
$$

Now we need to determine when such $z$ satisfies the condition $z^{2}=x y$, which, after substituting the found formula for $z$, takes the form

$$
\frac{x^{2} y^{2}(2 x+y)^{2}}{\left(x^{2}+2 y^{2}\right)^{2}}=x y
$$

After dividing by the number $x y \neq 0$ and removing the fraction, we get the condition

$$
x y(2 x+y)^{2}=\left(x^{2}+2 y^{2}\right)^{2} \quad \text { or } \quad(4 y-x)\left(x^{3}-y^{3}\right)=0 .
$$

The last equality holds precisely when either $x=4 y$ or $x^{3}=y^{3}$, i.e., $x=y \cdot{ }^{1}$ After substituting into the formula for $z$, we get $z=2 y$ in the first case, and $z=x$ in the second. Case (ii) thus corresponds to the solution groups $(4 t, t, 2 t)$ and $(t, t, t)$, where $t$ is any real number.

Answer. All solutions to the given system are $(t, 0,0),(0, t, 0),(0,0, t),(t, t, t)$, $(4 t, t, 2 t),(2 t, 4 t, t)$ and $(t, 2 t, 4 t)$, where $t$ is any real number.

Note. Let's show a way to avoid the more complex case (ii) in the given solution. Due to the cyclic symmetry, the given system implies not only the first of the following three equations (which we derived above), but also the other two analogous equations

$$
(2 x-y)\left(z^{2}-x y\right)=0, \quad(2 y-z)\left(x^{2}-y z\right)=0, \quad(2 z-x)\left(y^{2}-z x\right)=0 \text {. }
$$

We have analyzed the case $2 x-y=0$ above, and the cases $2 y-z=0$ and $2 z-x=0$ are analogous. By listing the solutions for these three cases, we obtain all the solutions mentioned in the Answer, except for $(t, t, t)$. If none of these three cases occur, the equations

$$
z^{2}-x y=x^{2}-y z=y^{2}-z x=0
$$

must be satisfied. Let's show that in the real number domain, they are satisfied only by the triples $(x, y, z)=(t, t, t)$. This is a simple consequence of the algebraic identity

$$
(x-y)^{2}+(y-z)^{2}+(z-x)^{2}=2\left(z^{2}-x y\right)+2\left(x^{2}-y z\right)+2\left(y^{2}-z x\right),
$$

whose right-hand side is zero according to (2), so the base of each of the three squares on the left-hand side (which would otherwise have a positive value) must also be zero. Let's add that the system (2) can also be solved by a shorter argument: if the equations (2) hold, the expressions $x^{3}, y^{3}, z^{3}$ have the same value $x y z$, and thus $x=y=z$ according to the note below the line.[^0]

Another solution. We will not repeat the initial consideration of the original solution and will directly look for only those solutions that satisfy the condition $x y z \neq 0$.

After dividing the expressions in the given system by the non-zero number $x y z$, we get

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

which is the equality of three values of the same function $f(s)=s+2 / s$ at non-zero points $s_{1}=y / z$, $s_{2}=z / x$ and $s_{3}=x / y$. We will therefore determine when for non-zero numbers $s$ and $t$ it holds that $f(s)=f(t)$. From the expression

$$
f(s)-f(t)=s+\frac{2}{s}-t-\frac{2}{t}=\frac{(s-t)(s t-2)}{s t}
$$

we see that the desired situation occurs only when $s=t$ or $s t=2$.

The system of equations (3) is therefore satisfied precisely when for the numbers $s_{1}, s_{2}, s_{3}$ it holds that: any two of them are equal or their product is 2. However, if such a product is 2, the third number is equal to $\frac{1}{2}$ due to the equality $s_{1} s_{2} s_{3}=1$, and the first two numbers (with a product of 2) thus lie in the set $\left\{\frac{1}{2}, 4\right\}$, so they are different, and therefore $\left(s_{1}, s_{2}, s_{3}\right)$ is some permutation of the triple $\left(\frac{1}{2}, \frac{1}{2}, 4\right)$. It is easy to verify that these (a total of three) permutations correspond to the solutions $(4 t, t, 2 t),(2 t, 4 t, t)$ and $(t, 2 t, 4 t)$ of the original system (we will not elaborate on this here). It is even easier to complete the remaining case $s_{1}=s_{2}=s_{3}$: due to the equality $s_{1} s_{2} s_{3}=1$, the common value of the numbers $s_{i}$ is 1 and the corresponding solutions are clearly $(t, t, t)$ (again, $t$ is any real number).

For a complete solution, award 6 points, of which 2 points for deriving the product form of at least one of the equations (1) or for transforming the system into the form (3) supplemented by a mention of the connection with the function $f(s)=s+2 / s$.