$$\Leftrightarrow 64(\Sigma ab)^3\ge 27(\Pi(a+b))^2$$ $$\\\Leftrightarrow 64\Sigma a^3b^3+192\Sigma a^3b^2c+192\Sigma a^2b^3c+384a^2b^2c^2\ge 27\Sigma (a^4b^2+a^2b^4)+54\Sigma a^4bc $$

由Muirhead Th. ,得证