26th Hellenic Mathematical Olympiad

We will use the well-known inequality $2\alpha \beta \le {\alpha }^2+{\beta }^2$, which is valid for all $\alpha ,\beta \in \mathbb{R}$. The equality holds for $\alpha =\beta$. Thus we have

$$\begin{array}{rl}{x}^2{y}^2+y^2{z}^2+z^2{x}^2+xyz& =\frac{1}{2}(2x^2{y}^2+2y^2{z}^2+2z^2{x}^2+2xyz)\\ & =\frac{1}{2}(2xy\cdot xy+2yz\cdot yz+2zx\cdot zx+2xyz)\\ & \le \frac{1}{2}[xy(x^2+y^2)+yz(y^2+z^2)+zx(z^2+x^2)+2xyz]\\ & =\frac{1}{2}[(xy+yz+zx)(x^2+y^2+z^2)-xy{z}^2-yz{x}^2-zx{y}^2+2xyz]\\ & =\frac{1}{2}[(xy+yz+zx)(x^2+y^2+z^2)-xyz(x+y+z-2)]\\ & =\frac{1}{2}[(xy+yz+zx)(x^2+y^2+z^2)],(\text{since }x+y+z=2).\end{array}\text{\hspace{1em}(1)}$$ Till now we have shown that $$x^2{y}^2+y^2{z}^2+z^2{x}^2+xyz\le \frac{1}{2}\left[(xy+yz+zx)(x^2+y^2+z^2)\right],(2)$$ and the equality holds, as we see from (1), when: $x=y=z$ or $x=y$, $z=0$ or $y=z$, $x=0$ or $z=x$, $y=0$. Since $x+y+z=2$, equality holds when: $$(x,y,z)=\left(\frac{2}{3},\frac{2}{3},\frac{2}{3}\right)\text{or}(1,1,0)\text{or}(1,0,1)\text{or}(0,1,1).(3)$$ In the sequel we will use the known inequality $\alpha \beta \le {\left(\frac{\alpha +\beta }{2}\right)}^2$, $\alpha ,\beta \in \mathbb{R}$, putting $\alpha =2xy+2yz+2zx$, $\beta =x^2+y^2+z^2$. Thus we have $$\begin{array}{rl}\frac{1}{2}\left[(xy+yz+zx)(x^2+y^2+z^2)\right]& =\frac{1}{4}\left[(2xy+2yz+2zx)(x^2+y^2+z^2)\right]\\ & \le \frac{1}{4}{\left(\frac{2xy+2yz+2zx+x^2+y^2+z^2}{2}\right)}^2=\frac{1}{16}(x+y+z{)}^4=1.\end{array}\text{\hspace{1em}(4)}$$ From (2) and (4) we obtain the inequality $x^2{y}^2+y^2{z}^2+z^2{x}^2+xyz\le 1.$ The equality holds when in inequality (4) we have: $\alpha =\beta \iff 2xy+2yz+2zx=x^2+y^2+z^2,$ which in consideration with (3) gives: $$(x,y,z)=(1,1,0)\text{or}(1,0,1)\text{or}(0,1,1).$$