Let $a=\tan{\alpha}$, $b=\tan{\beta}$ and $c=\tan{\gamma}$. then we have $$\frac{a^2}{1+a^2}=\sin^2{\alpha}, \frac{b^2}{1+b^2}=\sin^2{\beta}, \frac{c^2}{1+c^2}=\sin^2{\gamma}$$

Hence, we have $$\sin^2{\alpha}+\sin^2{\beta}+\sin^2{\gamma}=1$$

It follows that $$\cos^2{\alpha} = 1-\sin^2{\alpha} = sin^2{\beta} + \sin^2{\gamma}\ge 2\sin^2{\beta}\sin^2{\gamma}$$

Similarly, we shoul have $$\cos^2{\beta} \ge 2\sin{\alpha}\sin{\gamma}, \text{and}, \cos^2{\gamma} \ge 2\sin{\alpha}\sin{\beta}$$

Multiplying them together leads to $$\cos^{\alpha}\cos^{\beta}\cos^2{\gamma}\ge 8\sin^2{\alpha}\sin^2{\beta}\sin^2{\gamma}\implies \tan^2{\alpha} \tan^2{\beta} \tan^2{\gamma}\le\frac{1}{8}$$

Note that  $a=\tan{\alpha}$, $b=\tan{\beta}$ and $c=\tan{\gamma}$, then conclusion follows, i.e. $$abc\le\frac{\sqrt{2}}{4}$$