My Eleventh Problem

Just simple AM-GM will do: $\frac{4ab+6bc+8ac}{3}\geq\sqrt[3]{(4ab)(6bc)(8ac)}$ $\frac{9}{3}\geq\sqrt[3]{192(abc)^2}$ $3\geq\sqrt[3]{192(abc)^2}$ $27\geq192(abc)^2$ $\frac{27}{192}\geq(abc)^2$ $\frac{3}{8}\geq abc$

so the maximum value of $abc$ is $\boxed{0.375}$

Overrated! should have been a level 3 or 2 problem

simply apply am gm in the question the maximum value of abc comes out to be 3/8