Team Selection Test for JBMO 2024

$(x,y,z)=(\sqrt{6},\sqrt{6},\sqrt{6})$. Since these two triples are permutations of each other, sum of the elements of the first triple must be equal to sum of the elements of the second triple. Therefore we obtain $2(x^5+y^5+z^5)=72(x+y+z)$ and dividing by 2 we get $x^5+y^5+z^5=36(x+y+z)$. Combining this equation with the Power Mean Inequality we find $$12(x+y+z)=\frac{x^5+y^5+z^5}{3}\ge {\left(\frac{x+y+z}{3}\right)}^5$$ hence $x+y+z\le 3\sqrt{6}$. Combining this equality with the AM-GM inequality we get $$xyz\le {\left(\frac{x+y+z}{3}\right)}^3\le 6\sqrt{6}(1)$$

Now, we recall the equation $x^5+y^5+z^5=36(x+y+z)$. If we subtract the elements of the triple $(23x+24y+25z,23y+24z+25x,23z+24x+25y)$ from $36(x+y+z)$ we get the triple $(13x+12y+11z,13y+12z+11x,13z+12x+11y)$. Also, if we subtract the elements of the triple $(x^5+y^5,y^5+z^5,z^5+x^5)$ from $x^5+y^5+z^5$ we get the triple $(x^5,y^5,z^5)$. Since the initial triples are permutations of each other and the numbers we subtracted from are equal, we get that $(x^5,y^5,z^5)$ and $(13x+12y+11z,13y+12z+11x,13z+12x+11y)$ are permutations of each other as well. Hence, the product of the elements of the first triple must be equal to the product of the elements of the second triple: $$(xyz{)}^5=(13x+12y+11z)(13y+12z+11x)(13z+12x+11y)(2)$$

From the AM-GM inequality we have $$\sqrt{13x+12y+11z}\ge 36\sqrt[36]{x^{13}{y}^{12}{z}^{11}}$$ similarly we find $$13y+12z+11x\ge 36\sqrt[36]{y^{13}{z}^{12}{x}^{11}}$$ and $$13z+12x+11y\ge 36\sqrt[36]{z^{13}{x}^{12}{y}^{11}}$$ Multiplying all these inequalities and using it in (2) we find $$\begin{gathered} (xyz{)}^5=(13x+12y+11z)(13y+12z+11x)(13z+12x+11y)\ge 6^{6}xyz \\ \text{hence }xyz\ge 6\sqrt{6}.\text{ Combining this with (1), we get that }xyz=6\sqrt{6}. \\ \text{Therefore, all inequalities above should be equalities and hence} \\ x=y=z.\text{ We are done.} \end{gathered}$$