59th Ukrainian National Mathematical Olympiad

From the Arithmetic mean - Geometric mean Inequality, $$\begin{gathered} \frac{1}{2019}(a_1+a_2+\cdots +a_{2019})\ge \sqrt[2019]{a_1{a}_2\dots a_{2019}},\text{ or} \\ (a_1+a_2+\cdots +a_{2019}{)}^{2019}\ge 2019^{2019}{a}_1{a}_2\dots a_{2019}. \end{gathered}$$ From problem statement, $$\begin{gathered} (a_1+a_2+\cdots +a_{2019}{)}^{2019}=(\sqrt[2018]{2019^{2019}}{)}^{2019}=(2019\cdot \sqrt[2018]{2019}{)}^{2019}= \\ =2019^{2019}\cdot \sqrt[2018]{2019^{2019}}=2019^{2019}{a}_1{a}_2\dots a_{2019}, \end{gathered}$$ Which means that in this Arithmetic mean - Geometric mean Inequality, equality holds, which is possible iff $$a_1=a_2=\cdots =a_{2019}=\frac{1}{\sqrt[2019]{2019\cdot 2019^{2019}}}=\sqrt[2019]{2019^{2019}}.$$