China Western Mathematical Olympiad

Since $\sum a^3=1-3\Pi (a+b)$, $\sum a^5=1-5\Pi (a+b)[\sum a^2+\sum ab]$, therefore, the original inequality holds $$\begin{array}{rl}& \iff 10[1-3\Pi (a+b)]-9[1-5\Pi (a+b)(\sum a^2+\sum ab)]\ge 1\\ & \iff 45\Pi (a+b)(\sum a^2+\sum ab)\ge 30\Pi (a+b)\\ & \iff 3(\sum a^2+\sum ab)\ge 2=2(\sum a{)}^2=2(\sum a^2+2\sum ab)\\ & \iff \sum a^2\ge \sum ab.\end{array}$$ From $a^2+b^2\ge 2ab$, $b^2+c^2\ge 2bc$ and $c^2+a^2\ge 2ac$, we have $2\sum a^2\ge 2\sum ab$, i.e., $\sum a^2\ge \sum ab$. Therefore the original inequality holds.