11.4. Real numbers $x, y, z$ are such that $x+y+z=2$ and $xy+yz+zx=1$. Find the greatest possible value of the quantity $x-y$.
Answer: $\frac{2 \sqrt{3}}{3}$.

Solution: Eliminate the variable $z$:

$$
1=x y+z(x+y)=x y+(2-x-y)(x+y)=2 x+2 y-x^{2}-y^{2}-x y
$$

from which

$$
x^{2}+y^{2}+x y+1=2 x+2 y
$$

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

$$
a^{2}-\frac{a^{2}-b^{2}}{4}+1=2 a \Leftrightarrow 3 a^{2}+b^{2}+4=8 a \Leftrightarrow b^{2}=-3 a^{2}+8 a-4
$$

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

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

From this, it follows that $b^{2} \leq \frac{4}{3}$ and $x-y=b \leq \frac{2 \sqrt{3}}{3}$. Also note that the value $x-y=\frac{2 \sqrt{3}}{3}$ is achieved at $(x, y, z)=\left(\frac{2+\sqrt{3}}{3} ; \frac{2-\sqrt{3}}{3} ; \frac{2}{3}\right)$. 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.

Comment: A correct and justified solution - 7 points. The following criteria are cumulative.

![](https://cdn.mathpix.com/cropped/2024_05_06_e544875bf17d3fe51ac0g-3.jpg?height=86&width=1884&top_left_y=348&top_left_x=86)
A solution that meets the conditions of the problem, for which $x-y=\frac{2 \sqrt{3}}{3}-2$ points. Only the answer is provided - 0 points. The problem is not solved or solved incorrectly - 0 points.