Korean Mathematical Olympiad

Let $f(x,y,z)=(x^2-yz)(y^2-zx)(z^2-xy)$. Considering $f(x,0,z)=-x^3{z}^3$, it is easily seen that maximum of $f$ is positive. Since $f$ is symmetric in $x,y,z$ and $f(x,y,z)=f(-x,-y,-z)$, we may assume that $x\ge y\ge z$, $x+y+z\ge 0$ and consequently $x^2-yz>0$. If $f$ takes its maximum at $(x,y,z)$, then $$f(x,y,z)-f(x,-y,-z)=-2x(x^2-yz)(y^3+z^3)\ge 0.$$ Thus we have $y^3+z^3\le 0$. If $z=0$, then $y$ is also $0$ and thus $f(x,y,z)=0$. Therefore we must have $z<0$ and thus $y^2-zx>0$. Since $f(x,y,z)$ is positive $z^2-xy$ is also positive. Now applying the inequality of arithmetic and geometric means, we have $$f(x,y,z{)}^{\frac{1}{3}}\le \frac{(x^2-yz)+(y^2-zx)+(z^2-xy)}{3}=\frac{\frac{3}{2}-\frac{1}{2}(x+y+z{)}^2}{3}\le \frac{1}{2}$$ Consequently, $f$ takes its maximum $\frac{1}{3}$, in the case where $x+y+z=0$.