Since $ab+bc+ca\le \frac{(a+b+c)^2}{3}=3$ and $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le a+b+c=3$, by Cauchy-Schwarz we get: $\frac{\sqrt a +\sqrt b +\sqrt c}{ab+bc+ca} + \frac{1}{1+2\sqrt {ab}} + \frac {1}{1+ 2\sqrt {bc}} + \frac{1}{1+ 2\sqrt {ca}}\ge \frac{\sum_{cyc}\sqrt{a}}{3} +\frac{9}{2(\sum_{cyc}\sqrt{ab})+3}$ $(1)$. Note that $a+b+c=3\iff (\sum_{cyc}\sqrt{a})^2=2(\sum_{cyc}\sqrt{ab})+3$ then $RHS(1)$ $=\frac{\sqrt{2(\sum_{cyc}\sqrt{ab})+3}}{3}+\frac{9}{2(\sum_{cyc}\sqrt{ab})+3} =\frac{\sqrt{2(\sum_{cyc}\sqrt{ab})+3}}{6}+\frac{\sqrt{2(\sum_{cyc}\sqrt{ab})+3}}{6}+\frac{9}{2(2(\sum_{cyc}\sqrt{ab})+3)}+\frac{9}{2(2(\sum_{cyc}\sqrt{ab})+3)} \ge 3.\sqrt[3]{\frac{1}{6}.\frac{1}{6}.\frac{9}{2}}+\frac{9}{2.3+3}=\frac{3}{2}+\frac{1}{2}=2$. The equality occurs $\iff a=b=c=1$.

Is this in AoPS V2?