solved here

frengo wrote: for AM-GM, we have $a+b+c+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}+\frac{1}{9abc}$ $\geq12\sqrt[12]{\frac{1}{3^{18}a^8b^8c^8}}$ $=\frac{4}{\sqrt{3}\sqrt[3]{a^2b^2c^2}}$ it remains only to show that $\frac{4}{\sqrt{3}\sqrt[3]{a^2b^2c^2}}\geq4\sqrt{3}$ $\frac{1}{3}\geq \sqrt[3]{a^2b^2c^2}$ $\frac{a^2+b^2+c^2}{3}\geq \sqrt[3]{a^2b^2c^2}$ true for AM-GM.