International Zhautykov Olympiad

Let $d_{1}=A B \cdot B C \cdot C D \cdot D E \cdot E F \cdot F A, d_{2}=A C \cdot B D \cdot C E \cdot D F \cdot A E \cdot B F, d_{3}=A D \cdot B E \cdot C F$. Applying Ptolemy's theorem to quadrilaterals $A B C D, B C D E, C D E F, D E F A, E F A B, F A B C$, we obtain six equations $A C \cdot B D-A B \cdot C D=B C \cdot A D, \ldots, F B \cdot A C-F A \cdot B C=A B \cdot F C$. Putting these equations in the well-known inequality $$\sqrt[6]{\left(a_{1}-b_{1}\right)\left(a_{2}-b_{2}\right) \cdot \ldots \cdot\left(a_{6}-b_{6}\right)} \leq \sqrt[6]{a_{1} a_{2} \ldots a_{6}}-\sqrt[6]{b_{1} b_{2} \ldots b_{6}} \quad\left(a_{i} \geq b_{i}>0, i=1, \ldots, 6\right)$$ we get $$\sqrt[3]{d_{3}} \sqrt[6]{d_{1}} \leq \sqrt[3]{d_{2}}-\sqrt[3]{d_{1}} \tag{1}$$ Applying Ptolemy's theorem to quadrilaterals $A C D F, A B D E$ и $B C E F$, we obtain three equations $A D \cdot C F=A C \cdot D F+A F \cdot C D, A D \cdot B E=B D \cdot A E+A B \cdot D E, B E \cdot C F=B F \cdot C E+B C \cdot E F$. Putting these equations in the well-known inequality $$\sqrt[3]{\left(a_{1}+b_{1}\right)\left(a_{2}+b_{2}\right)\left(a_{3}+b_{3}\right)} \geq \sqrt[3]{a_{1} a_{2} a_{3}}+\sqrt[3]{b_{1} b_{2} b_{3}}\left(a_{i}>0, b_{i}>0, i=1,2,3\right)$$ we get $$\sqrt[3]{d_{3}^{2}} \geq \sqrt[3]{d_{2}}+\sqrt[3]{d_{1}} \tag{2}$$ It follows from (1) and (2) that $(\sqrt[3]{d_{2}}-\sqrt[3]{d_{1}})^{2} \geq \sqrt[3]{d_{1}}(\sqrt[3]{d_{2}}+\sqrt[3]{d_{1}})$, that is, $\sqrt[3]{d_{2}} \geq 3 \sqrt[3]{d_{1}}$ and $d_{2} \geq 27 d_{1}$, q.e.d.