(Turkish Junior Olympiads 2021)

Let $x, y,$ and $z$ be three real numbers satisfying the equations $x+y+z=2$ and $xy+yz+zx=1$. Find the maximum value that $x-y$ can take.
We start by combining the two equations to eliminate one of the three variables (here $z$ since it does not appear in $x-y$). We have $1=x y+z(x+y)=x y+(2-x-y)(x+y)$, which gives the equation

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

We introduce the reduced variables $a=x-y$ (which is what we are looking for) and $b=x+y$, with which we can rewrite this relation more simply, which becomes $2 b=b^{2}-\frac{b^{2}-a^{2}}{4}+1$. Thus, $a^{2}=$ $-3 b^{2}+8 b-4$

It only remains to find the maximum value reached by the function $f: x \rightarrow-3 x^{2}+8 x-4$, which is reached at $\frac{-8}{2 \times-3}=\frac{4}{3}$. We have $f\left(\frac{4}{3}\right)=\frac{4}{3}$, hence $a^{2} \leq \frac{4}{3}$ and $\frac{-2}{\sqrt{3}} \leq x-y \leq \frac{2}{\sqrt{3}}$

## Algebra