Macedonian Mathematical Olympiad

If we use the Cauchy-Bunyakovsky-Schwarz inequality for the positive numbers $(x,y^2,z^3)$ and $(x^3,y^2,z)$ we get $$(x^4+y^4+z^4{)}^2\le (x^2+y^4+z^6)(x^6+y^4+z^2)\text{ i.e. }$$ $$\frac{1}{x^2+y^4+z^6}\le \frac{x^6+y^4+z^2}{9}.$$ (1) Analogously, using the Cauchy-Bunyakovsky-Schwarz inequality for the positive numbers $(x^2,y^3,z)$ and $(x^2,y,z^3)$, and also for the positive numbers $(x^3,y,z^2)$ and $(x,y^3,z^2)$, we get $$\frac{1}{x^4+y^6+z^2}\le \frac{x^4+y^2+z^6}{9},$$ (2) i.e. $$\frac{1}{x^6+y^2+z^4}\le \frac{x^2+y^6+z^4}{9}.$$ (3) Now, by adding the inequalities (1), (2) and (3) we get the inequality $$\frac{1}{x^2+y^4+z^6}+\frac{1}{x^4+y^6+z^2}+\frac{1}{x^6+y^2+z^4}\le \frac{x^6+y^6+z^6+x^4+y^4+z^4+x^2+y^2+z^2}{9}.$$ (4)

From the inequality between the arithmetic and quadratic mean for the positive numbers $x^2,y^2$ and $z^2$ we get that $x^2+y^2+z^2\le 3\sqrt{\frac{x^4+y^4+z^4}{3}}=3$, so if we substitute in (4) we get the required inequality. Equality in (1) holds if and only if $\frac{x}{x^3}=\frac{y^2}{y^2}=\frac{z^3}{z^3}$ i.e. $x=z=1$ and from $x^4+y^4+z^4=3$ it follows that $y=1$.