40th Hellenic Mathematical Olympiad

1. (First solution) The given equations can be written as: $$x{y}^{2}z+xyz+y=xyzw+yzw+x$$ $$z^{2}yw+yzw+z=xyzw+y+zwx$$ $$w^{2}xz+zwx+w=xyzw+wxy+z.$$ $$x^{2}wy+wxy+x=xyzw+xyz+w.$$ Summing up the four equations we get: $$x{y}^{2}z+x^{2}wy+y{z}^{2}w+w^{2}xy=4xyzw\iff \frac{y}{w}+\frac{x}{z}+\frac{z}{x}+\frac{w}{y}=4.$$ From arithmetic-geometric means inequality we have: $$\frac{y}{w}+\frac{x}{z}+\frac{z}{x}+\frac{w}{y}\ge 4.$$ Where the equality holds, if and only if $$\frac{y}{w}=\frac{x}{z}=\frac{z}{x}=\frac{w}{y}\iff x^2=z^2\text{ and }{y}^2=w^2\iff x=z\text{ and }y=w.$$

$$\begin{gathered} \frac{xyz+1}{x+1}=\frac{yzw+1}{y+1}\iff \frac{x^{2}y+1}{x+1}=\frac{y^{2}x+1}{y+1}=\frac{x^{2}y-y^{2}x}{x-y}=xy \\ \iff x^2{y}^2+x^{2}y+1+y=x^2{y}^2+y^{2}x+1+x \end{gathered}$$ $$\iff xy(x-y)=x-y\iff (x-y)(xy-1)=0\iff x=y\text{, }xy=1.$$ We distinguish two cases: 1. $x=y=z=w$. Then, from $x+y+z+w=48$, we have the solution $(x,y,z,w)=(12,12,12,12)$. 2. $x=z$ and $y=w$ and $xy=1$. Then $$x+y=24\iff x+\frac{1}{x}=24\iff x^2-24x+1=0\iff x=12\pm \sqrt{143},$$ and we find the solutions $$(x,y,z,w)=(12+\sqrt{143},12-\sqrt{143},12+\sqrt{143},12-\sqrt{143})$$ $$(x,y,z,w)=(12-\sqrt{143},12+\sqrt{143},12-\sqrt{143},12+\sqrt{143}).$$