28th Hellenic Mathematical Olympiad

Since $x$, $y$, $z$ are positive integers with sum $12$, it is enough to prove that $$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+\frac{x+y+z}{4}\ge \sqrt{x}+\sqrt{y}+\sqrt{z}.(1)$$ From the inequality of the arithmetic–geometric mean for the positive integers $x$, $y$, $z$ we get

$$\frac{x}{y}+\frac{y}{4}\ge 2\sqrt{\frac{x}{y}\cdot \frac{y}{4}}=\sqrt{x},(2)$$ $$\frac{y}{z}+\frac{z}{4}\ge 2\sqrt{\frac{y}{z}\cdot \frac{z}{4}}=\sqrt{y},(3)$$ $$\frac{z}{x}+\frac{x}{4}\ge 2\sqrt{\frac{z}{x}\cdot \frac{x}{4}}=\sqrt{z}.(4)$$ Summing up (2), (3) and (4) we find inequality (1).

Equality holds when all inequalities (2), (3) and (4) hold as equalities, that is when $$\begin{array}{rl}\frac{x}{y}=\frac{y}{4},\frac{y}{z}=\frac{z}{4},\frac{z}{x}=\frac{x}{4}& \iff x=\frac{y^2}{4},y=\frac{z^2}{4},z=\frac{1}{4}{\left(\frac{y^2}{4}\right)}^2=\frac{y^4}{4^3}\\ & \iff x=\frac{y^2}{4},y=\frac{z^2}{4},z=\frac{1}{4^3}{\left(\frac{z^2}{4}\right)}^4\\ & \iff x=\frac{y^2}{4},y=\frac{z^2}{4},z=\frac{1}{4^7}{z}^8\\ & \iff x=\frac{y^2}{4},y=\frac{z^2}{4},z^7=4^7\\ & \iff x=y=z=4.\end{array}$$