Find the minimum value of $$\sum \frac {\sqrt a}{\sqrt b +\sqrt c-\sqrt a}$$ Where $a, b, c$ represent the sides of a triangle
My approach
$$\sum \frac {\sqrt a}{\sqrt b +\sqrt c-\sqrt a}$$ $$=\sum \frac {1}{\sqrt {\frac ba} +\sqrt {\frac ca}-1}$$ By Titu's lemma
$$=\sum \frac {1}{\sqrt {\frac ba} +\sqrt {\frac ca}-1}\ge \frac {9}{\left ( \sum \sqrt {\frac ba}+ \sqrt {\frac ab}\right)-3}$$
Now by Cauchy Schwarz inequality $$\left ( \sum \sqrt {\frac ba}+ \sqrt {\frac ab}\right)^2 \le \left (\sum \frac ab +\frac ba\right)(6)$$ Hence we have $$=\sum \frac {1}{\sqrt {\frac ba} +\sqrt {\frac ca}-1}\ge \frac {9}{\left ( \sum \sqrt {\frac ba}+ \sqrt {\frac ab}\right)-3}\ge \frac {9}{\sqrt {\left (\sum \frac ab +\frac ba\right)(6)}-3}$$
Now how to proceed further.