⇔64(Σab)3≥27(Π(a+b))2\Leftrightarrow 64(\Sigma ab)^3\ge 27(\Pi(a+b))^2 ⇔64Σa3b3+192Σa3b2c+192Σa2b3c+384a2b2c2≥27Σ(a4b2+a2b4)+54Σa4bc\\\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

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