let $ p=\sqrt{\frac{1}{a}-1}$ and so on then since $ a+b+c=1$ we have $ p^2q^2+q^2r^2+r^2p^2+2=p^2q^2r^2$ so let $ pq=\frac{1}{\cos A}$ and so on. it suffices to prove $ \sum \frac{1}{\cos A} \ge 6$.

moldovan wrote: The sum of positive real numbers $ a,b,c$ equals $ 1$. Prove that: $ \sqrt {\frac {1}{a} - 1}\sqrt {\frac {1}{b} - 1} + \sqrt {\frac {1}{b} - 1}\sqrt {\frac {1}{c} - 1} + \sqrt {\frac {1}{c} - 1}\sqrt {\frac {1}{a} - 1} \ge 6.$ $ \sqrt {\frac {1}{a} - 1} = \sqrt {\frac {b + c}{a}} \ge \frac {\sqrt {2}\sqrt [4]{bc}}{\sqrt{a}}$. Using this everywhere, it becomes: $ \sum_{cyc} \sqrt [4]{\frac {c^2}{ab}} \ge 3$ which is obvious by AM-GM..