19th Turkish Mathematical Olympiad

By the Cauchy-Schwarz inequality we have $$\frac{1}{a+b^{20}+c^{11}}\le \frac{a^{13}+b^{-6}+c^3}{(a^7+b^7+c^7{)}^2}$$ for all positive real numbers $a,b,c$. Summing up the inequalities for $(a,b,c)=(x,y,z),(y,z,x)$ and $(z,x,y)$ gives that it suffices to show that $$x^{13}+y^{13}+z^{13}+x^{-6}+y^{-6}+z^{-6}+x^3+y^3+z^3\le x^{14}+y^{14}+z^{14}+2(x^7{y}^7+y^7{z}^7+z^7{x}^7)$$ to solve the problem. As $xyz=1$ we have $x^{13}+y^{13}+z^{13}={\sum }_{cyc}{x}^{13\frac{1}{3}}{y}^{\frac{1}{3}}{z}^{\frac{1}{3}}$, $x^{-6}+y^{-6}+z^{-6}={\sum }_{cyc}{x}^{6\frac{2}{3}}{y}^{6\frac{2}{3}}{z}^{\frac{2}{3}}$ and $x^3+y^3+z^3={\sum }_{cyc}{x}^{6\frac{2}{3}}{y}^{3\frac{2}{3}}{z}^{3\frac{2}{3}}$. Then by Muirhead's inequality we obtain $$\begin{gathered} \sum _{cyc}{x}^{13\frac{1}{3}}{y}^{\frac{1}{3}}{z}^{\frac{1}{3}}\le \sum _{cyc}{x}^{14}{y}^0{z}^0,\sum _{cyc}{x}^{6\frac{2}{3}}{y}^{6\frac{2}{3}}{z}^{\frac{2}{3}}\le \sum _{cyc}{x}^7{y}^7{z}^0\text{and}\sum _{cyc}{x}^{6\frac{2}{3}}{y}^{3\frac{2}{3}}{z}^{3\frac{2}{3}}\le \sum _{cyc}{x}^7{y}^7{z}^0 \\ \text{since }(13\frac{1}{3},\frac{1}{3},\frac{1}{3})\prec (14,0,0),(6\frac{2}{3},6\frac{2}{3},\frac{2}{3})\prec (7,7,0)\text{ and }(6\frac{2}{3},3\frac{2}{3},3\frac{2}{3})\prec (7,7,0). \end{gathered}$$