We have : $tan^2\alpha +1 =\frac{1}{1-sin^2\alpha}$ The problem is equivalent to : $\frac{1}{1-sin^2\alpha}+\frac{1}{1-sin^2\beta}+\frac{1}{1-sin^2\gamma} \geq \frac{27}{8}$ By Cauchy-Scharwz's ineq , we have : $\frac{1}{1-sin^2\alpha}+\frac{1}{1-sin^2\beta}+\frac{1}{1-sin^2\gamma} \geq \frac{9}{3-(sin^2\alpha+sin^2\beta+sin^2\gamma)}$ It's enough to show that : $sin^2\alpha+sin^2\beta+sin^2\gamma \geq \frac{1}{3}$ , which is true by Cauchy again .

by cauchy we have (cos²a+ cos²b+ cos²c) . (tg²a+ tg²b + tg²c) >= (sina +sinb +sinc)² = 1. see that, again by cauchy, 3.(sin²a + sin²b + sin²c) >= (sina + sinb+ sinc)² = 1 this implies that 3- (cos²a + cos²b + cos²c)>= 1/3 so, cos²a+cos²b+cos²c <=3/8. so, 3/8 . (tg²a+tg²b+tg²a) >= 1, and the result follow.

My solution is similar to nntu. put $sin a=x$ etc. So the problem is equivalent to. $x,y,z>0$ $x+y+z=1$ and we want to prove that $\sum\frac{x^2}{1-x^2}\geq\frac{3}{8}$. We use cauchy in the engel form to have $\sum\frac{x^2}{1-x^2}\geq\frac{(x+y+z)^2}{3-(x^2+y^2+z^2)}$ so it emains to prove that $\frac{(x+y+z)^2}{3-(x^2+y^2+z^2)}\geq\frac{3}{8}$ <=>...<=> $3(x^2+y^2+z^2)\geq 1$ which is true.