I don't have a nice proof, but it is a solution for the problem. At first, let $ a=BC,b=CA,c=AB $. And $ D,E $ be the intersections of the in-circle and $ AB,AC $, respectively. Then, we need to calculate the length of the segment $ l_{1} $. Notice that $ \triangle AMN~\triangle ABC $. Thus, \[ \frac{l_{1}}{t_{a}}=\frac{AM+MN+NA}{a+b+c}=\frac{AM+(MI+IN)+NA}{a+b+c}=\frac{(AM+MI)+(IN+NA)}{a+b+c}=\frac{AD+AE}{a+b+c}=\frac{b+c-a} {a+b+c} \] And use a well-known equality, that is, $ t_a=\frac{\sqrt{bc(b+c-a)(a+b+c)}}{b+c} $. We have \[ l_{1}=\frac{(b+c-a)\sqrt{bc(b+c-a)(a+b+c)}}{(b+c)(a+b+c)} \] Now, come back to the original question~ \[ \frac {1}{l^{2}_{1}}+\frac {1}{l^{2}_{2}}+\frac {1}{l^{2}_{3}} \ge \frac{81}{p^{2}} \Leftrightarrow \sum\limits_{cyc}{\frac{(b+c)^2(a+b+c)^3}{bc(b+c-a)^3}} \ge 324 \] It's easy by using AM-GM inequality. Precisely, \[ \sum\limits_{cyc}{\frac{(b+c)^2(a+b+c)^3}{bc(b+c-a)^3}} \ge 4\sum\limits_{cyc}{\frac{(a+b+c)^3}{(b+c-a)^3}} \ge 12\cdot\frac{(a+b+c)^3}{(b+c-a)(c+a-b)(a+b-c)} \ge 324 \]