Estonian Mathematical Olympiad

By bringing all the terms to the same side and to the common denominator, we get an equivalent inequality $$\frac{y^{2}z(x+y^2+z)+x{y}^2(x+y^2+z)+xz(x+y^2+z)-9x{y}^{2}z}{x(x+y^2+z)}\ge 0.$$ Since $x$, $y$, and $z$ are positive, the denominator $x(x+y^2+z)$ is positive as well. Therefore the fraction on the left-hand side is nonnegative if and only if its numerator is nonnegative. By removing the parentheses, we get an equivalent inequality $x^2{y}^2+x{y}^4+y^2{z}^2+y^{4}z+x^{2}z+x{z}^2⩾6x{y}^{2}z$. However, this inequality follows directly from AM-GM for terms $x^2{y}^2$, $x{y}^4$, $y^2{z}^2$, $y^{4}z$, $x^{2}z$, and $x{z}^2$.

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Alternative solution.

When multiplying both sides of the inequality by $\frac{x+z+y^2}{y^{2}z}$ we get an equivalent inequality $(\frac{y^{2}z}{x}+y^2+z)(\frac{x+z+y^2}{y^{2}z})⩾9$. This inequality can in turn be transformed into $(\frac{y^{2}z}{x}+y^2+z)(\frac{x}{y^{2}z}+\frac{1}{y^2}+\frac{1}{z})⩾9$. The last one, however, follows from Cauchy-Schwarz, when it is applied to $(\frac{y\sqrt{z}}{\sqrt{x}},y,\sqrt{z})$ and $(\frac{\sqrt{x}}{y\sqrt{z}},\frac{1}{y},\frac{1}{\sqrt{z}})$.