SELECTION and TRAINING SESSION

Write ${\sigma }_1=a+b+c$, ${\sigma }_2=ab+bc+ca$, ${\sigma }_3=abc$. Then the inequality we have to prove becomes $${\sigma }_1^5\ge 81({\sigma }_1^2-2{\sigma }_2){\sigma }_3⟺{\sigma }_1^5+162{\sigma }_2{\sigma }_3\ge 81{\sigma }_1^2{\sigma }_3.(1)$$

Now using the AM-GM inequality for three summands ${\sigma }_1^5$, $81{\sigma }_2{\sigma }_3$, $81{\sigma }_2{\sigma }_3$ we get $${\sigma }_1^5+81{\sigma }_2{\sigma }_3+81{\sigma }_2{\sigma }_3\ge 3\sqrt[3]{81^2{\sigma }_1^5{\sigma }_2^2{\sigma }_3^2}.$$ So for (1) to be proved it suffices to verify that $$3\sqrt[3]{81^2{\sigma }_1^5{\sigma }_2^2{\sigma }_3^2}\ge 81{\sigma }_1^2{\sigma }_3⟺81^2{\sigma }_1^5{\sigma }_2^2{\sigma }_3^2\ge 27^3{\sigma }_1^6{\sigma }_3^3⟺{\sigma }_2^2\ge 3{\sigma }_1{\sigma }_3.$$ The last inequality is well known.