I saw this question in one of my Trigonometry textbooks and I am unable to find any way to move ahead. The question says that let $A,B$ and $C$ be the angles of a triangle, then prove that
$$ \sum_{cyc} \frac{\sqrt { \sin A}}{\sqrt { \sin B}+\sqrt { \sin C}-\sqrt { \sin A}} \ge \sqrt3 $$
Kindly tell me about a good approach to strike the question.
We need to prove that $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}\geq\sqrt3.$$ Now, by C-S $$\sum_{cyc}\frac{\sqrt{a}}{\sqrt{b}+\sqrt{c}-\sqrt{a}}=\sum_{cyc}\frac{a}{\sqrt{ab}+\sqrt{ac}-a}\geq$$ $$\geq\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sum\limits_{cyc}(\sqrt{ab}+\sqrt{ac}-a)}=\frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{\sum\limits_{cyc}(2\sqrt{ab}-a)}.$$ Thus, it remains to prove that $$\sum_{cyc}(a+2\sqrt{ab})\geq\sqrt3\sum_{cyc}(2\sqrt{ab}-a)$$ or $$(1+\sqrt3)(a+b+c)\geq2(\sqrt3-1)(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}),$$ which is true because $1+\sqrt3>2(\sqrt3-1)$ and $a+b+c\geq\sqrt{ab}+\sqrt{ac}+\sqrt{bc}$.
Done!