We know that for $∆ABC$

$\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right)\tan \left(\frac{A}{2}\right)=1 ....\left(\mathrm{i}\right)$

For triangle, $\tan \left(\frac{A}{2}\right), \tan \left(\frac{B}{2}\right), \tan \left(\frac{C}{2}\right)>0$

Now,

$\mathrm{AM}\ge \mathrm{GM}$

$\Rightarrow \frac{\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)+\tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)+\tan \left(\frac{C}{2}\right)\tan \left(\frac{A}{2}\right)}{3}\ge {\left[\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)\times \tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)\times \tan \left(\frac{C}{2}\right)\tan \left(\frac{A}{2}\right)\right]}^{\frac{1}{3}}$

$\Rightarrow {\left[{\tan }^2\left(\frac{A}{2}\right){\tan }^2\left(\frac{B}{2}\right){\tan }^2\left(\frac{C}{2}\right)\right]}^{\frac{1}{3}}\le \frac{1}{3}$

$\Rightarrow {\left[\tan \left(\frac{A}{2}\right)\tan \left(\frac{B}{2}\right)\tan \left(\frac{C}{2}\right)\right]}^2\le \frac{1}{27}$