Problem 5. Real numbers $x, y, z$ are such that $x+y+z=2$ and $x y+y z+z x=1$. Find the greatest possible value of the quantity $x-y$.

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Answer: $\frac{2 \sqrt{3}}{3}$.

Solution. Eliminate the variable $z$:

$$
\begin{gathered}
1=x y+z(x+y)=x y+(2-x-y)(x+y)=2 x+2 y-x^{2}-y^{2}-x y \Rightarrow \\
x^{2}+y^{2}+x y+1=2 x+2 y .
\end{gathered}
$$

Let $a=x+y$ and $b=x-y$, express everything in terms of $a$ and $b$:

$$
\begin{gathered}
x^{2}+y^{2}+x y+1=2 x+2 y \Rightarrow a^{2}-\frac{a^{2}-b^{2}}{4}+1=2 a \Rightarrow 3 a^{2}+b^{2}+4=8 a \Rightarrow \\
b^{2}=-3 a^{2}+8 a-4
\end{gathered}
$$

Estimate the value of $b^{2}=-3 a^{2}+8 a-4$ from above:

$$
\left(-3 a^{2}+8 a-4\right)-\frac{4}{3}=-3\left(a-\frac{4}{3}\right)^{2} \leqslant 0 \Rightarrow-3 a^{2}+8 a-4 \leqslant \frac{4}{3}
$$

From this, it follows that $b^{2} \leqslant \frac{4}{3}$ and $x-y=b \leqslant \frac{2 \sqrt{3}}{3}$.

Also note that the value $x-y=\frac{2 \sqrt{3}}{3}$ is achieved when $(x, y, z)=\left(\frac{2+\sqrt{3}}{3}, \frac{2-\sqrt{3}}{3}, \frac{2}{3}\right)$. Indeed, in this case $a=\frac{4}{3}, b=\frac{2 \sqrt{3}}{3}, b^{2}=-3 a^{2}+8 a-4$, so all the inequalities given above turn into equalities.

Criteria

The following criteria are summed:

5 6. It is proven that $x-y \leqslant \frac{2 \sqrt{3}}{3}$.

2 6. It is proven that there exists a triplet of numbers $(x, y, z)$ satisfying the conditions of the problem for which $x-y=\frac{2 \sqrt{3}}{3}$.

Criteria

The largest suitable criterion is used:

7 points. Any complete solution to the problem.

2 points. There is only the correct answer.