Exercise 8. Let $x, y, z$ be three real numbers satisfying $x+y+z=2$ and $xy+yz+zx=1$. Determine the maximum value that $x-y$ can take.
## Solution to Exercise 8

Let $(x, y, z)$ be a triplet satisfying the statement. By swapping $x$ and the maximum of the triplet, and then $z$ and the minimum of the triplet, we can assume $x \geqslant z \geqslant y$, while increasing $x-y$.

Since $x+y+z=2$, we have $4=(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2(x y+y z+x z)=x^{2}+y^{2}+z^{2}+2$, thus $x^{2}+y^{2}+z^{2}=2$.

In particular, $(x-y)^{2}+(y-z)^{2}+(x-z)^{2}=2\left(x^{2}+y^{2}+z^{2}\right)-2(x y+y z+x z)=2$. By the arithmetic-quadratic inequality, $(x-z)^{2}+(z-y)^{2} \geqslant \frac{(x-z+z-y)^{2}}{2}=\frac{(x-y)^{2}}{2}$, so $2 \geqslant(x-y)^{2}+\frac{(x-y)^{2}}{2}=\frac{3(x-y)^{2}}{2}$, thus $(x-y) \leqslant \frac{2}{\sqrt{3}}$.

Let's try to show that this value is achievable. If we have $x-y=\frac{2}{\sqrt{3}}$, then we have equality in the arithmetic-quadratic inequality: thus $x-z=z-y$, so since their sum is $x-y$, we have $x-z=z-y=\frac{1}{\sqrt{3}}$. In particular, since $x+y+z=3 z+(z-x)+(z-y)=3 z$, we have $z=\frac{2}{3}, x=\frac{2}{3}+\frac{1}{\sqrt{3}}$ and $y=\frac{2}{3}-\frac{1}{\sqrt{3}}$. Conversely, for these values of $x, y, z$, we have:

$-x+y+z=3 \times \frac{2}{3}=2$

$-x y+y z+z x=x y+(x+y) z=\left(\frac{2}{3}+\frac{1}{\sqrt{3}}\right)\left(\frac{2}{3}-\frac{1}{\sqrt{3}}+\frac{4}{3} \times \frac{2}{3}=\frac{4}{9}-\frac{1}{3}+\frac{8}{9}=\frac{9}{9}=1\right.$

$-x-y=2 \times \frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$

Thus $(x, y, z)$ satisfies the conditions of the statement, and $x-y=\frac{2}{\sqrt{3}}$, so the maximum value that $x-y$ can take is $\frac{2}{\sqrt{3}}$.

Graders' Comment: The exercise was not approached by many. The main error is forgetting to find $x, y, z$ for which equality holds.