37th Iranian Mathematical Olympiad

We know that ${\sum }_{cyc}{a}^2+{\sum }_{cyc}ab={\sum }_{cyc}a+3abc$, so $$\begin{array}{rl}{\left(\sum _{cyc}{a}^2+\sum _{cyc}ab\right)}^2& ={\left(\sum _{cyc}a+3abc\right)}^2\\ \Rightarrow & {\left(\sum _{cyc}{a}^2\right)}^2+{\left(\sum _{cyc}ab\right)}^2+2\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right)\\ & ={\left(\sum _{cyc}a\right)}^2+9a^2{b}^2{c}^2+6abc\sum _{cyc}a\\ \Rightarrow & {\left(\sum _{cyc}{a}^2\right)}^2+{\left(\sum _{cyc}ab\right)}^2-{\left(\sum _{cyc}a\right)}^2-9a^2{b}^2{c}^2\\ & =6abc\sum _{cyc}a-2\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right).\end{array}$$ By AM-GM we have $$\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right)\ge {\left(\sum _{cyc}ab\right)}^2\ge 3abc\sum _{cyc}a$$ So $$2\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right)-6abc\sum _{cyc}a\ge 6abc\sum _{cyc}a-2\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right)$$

Then using above facts $$\begin{array}{rl}& 2\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right)-6abc\sum _{cyc}a\\ & \ge {\left(\sum _{cyc}{a}^2\right)}^2+{\left(\sum _{cyc}ab\right)}^2-{\left(\sum _{cyc}a\right)}^2-9a^2{b}^2{c}^2\\ \iff & {\left(\sum _{cyc}a\right)}^2+9a^2{b}^2{c}^2-6abc\sum _{cyc}a\\ & \ge {\left(\sum _{cyc}{a}^2\right)}^2-2\left(\sum _{cyc}{a}^2\right)\left(\sum _{cyc}ab\right)+{\left(\sum _{cyc}ab\right)}^2\\ \iff & {\left(\sum _{cyc}a-3abc\right)}^2\ge {\left(\sum _{cyc}{a}^2-\sum _{cyc}ab\right)}^2.\end{array}$$ Since ${\sum }_{cyc}{a}^2\ge {\sum }_{cyc}ab$, we have ${\sum }_{cyc}{a}^2-{\sum }_{cyc}ab\ge 0$. Therefore $$\begin{array}{rl}\left|\sum _{cyc}a-3abc\right|& \ge \sum _{cyc}{a}^2-\sum _{cyc}ab\\ \iff & \left|2\sum _{cyc}a-6abc\right|\ge \sum _{cyc}(a-b{)}^2.\end{array}$$