75th Romanian Mathematical Olympiad

If $x\ge 0$, $y\ge 0$, and $z\ge 0$, then $ma-nb=c(m-n)$ and analogously, we have $$m(a-c)=n(b-c),m(b-a)=n(c-a),m(c-b)=n(a-b).(\ast )$$ If all parentheses in $(\ast )$ are nonzero, then we obtain $m^3(a-c)(b-a)(c-b)=-n^3(b-c)(c-a)(a-b)$, and therefore $m=-n$, false. Thus at least one difference is zero, say $a-c=0$. It follows that $b=c$, hence $a=b=c$.