$\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}+\frac{16}{abc}+\frac{16}{abc}}{5}>=(\frac{256}{abc})^{\frac{1}{5}}>=2$ (AM-GM)

$\frac{ab+bc+ac}{3}>=\sqrt[3]{a^2b^2c^2}$

$64>=a^2b^2c^2$

$8>=abc$

$\frac{1}{abc}>=\frac{1}{8}$