Ukrajina 2008

Let $B=\frac{1}{x-y}+\frac{1}{y-z}+\frac{1}{z-x}=\frac{3}{2}$ and $A=\frac{1}{(x-y{)}^2}+\frac{1}{(y-z{)}^2}+\frac{1}{(z-x{)}^2}$ then $$B^2=\frac{9}{4}=A+2\left(\frac{1}{(x-y)(y-z)}+\frac{1}{(x-y)(z-x)}+\frac{1}{(y-z)(z-x)}\right).$$ Let's rearrange the second summand of the last expression: $$\frac{1}{(x-y)(y-z)}+\frac{1}{(x-y)(z-x)}+\frac{1}{(y-z)(z-x)}=\frac{(z-x)+(y-z)+(x-y)}{(x-y)(y-z)(z-x)}=0.$$ Therefore, $A=\frac{9}{4}$.

Though it is not required in the problem, we should check that there are numbers $x$, $y$, $z$ which satisfy the given condition. Let $x=2$, $y=1$, $z=0$ then $(x-y)=1$, $(y-z)=1$, $(z-x)=-2$, $B=1+1-\frac{1}{2}=\frac{3}{2}$, and $A=1+1+\frac{1}{4}=\frac{9}{4}$.