I think I recall something about [ABC] and the product of the sines of the angles, but I don't remember what it was. Feel free to correct me on that matter.

this problem is much easier than that, the inequality is equivalent to $ \sum\dfrac{1}{\sqrt a}\leq \sqrt {\dfrac{s}{Rr}} = \dfrac{2s}{\sqrt {abc}}$. after multiplying it follows that the inequality is equivalent to $ \sum\sqrt {bc}\leq 2s$, which is well-known... (i mean, it follows from the well-known fact $ x^2+y^2+z^2\geq xy+yz+zx$)

$ \sum_{} {\frac {1}{\sqrt {2sinA}} \leq {\sqrt \frac {p}{r}}}$ $ \Leftrightarrow$ $ \sum_{} {\sqrt {\frac {r}{2sinA}}} \leq \sqrt p$ $ \Leftrightarrow$ $ \sum_{} {\sqrt {\frac {bc}{2(a + b + c)}}} \leq \sqrt {\frac {(a + b + c)}{2}}$ $ \Leftrightarrow$ $ \sum_{} {\sqrt {bc}} \leq a + b + c$ $ \Leftrightarrow$ $ \sum_{} {{(\sqrt a - \sqrt b)}^2} \geq 0$ QED

We know that $ S = \frac {1}{2}ab\sin C \Longrightarrow 2\sin C = \frac{4S}{ab} $ from here follows that: $ \sqrt {\frac{ab}{4S}} + \sqrt{\frac{bc}{4S}} + \sqrt{\frac{ac}{4S}}$ $=$ $\frac{\sqrt{ab} + \sqrt{bc} +\sqrt{ac}}{\sqrt{4p\cdot r}}$ $\sqrt{ab} \leq \frac{a+b}{2}$ ,$\sqrt{bc} \leq \frac{b+c}{2}$ and $\sqrt{ac} \leq \frac{a+c}{2}$ , because $ a, b, c $ are the sides of triangle. Now adding these 3 relations we see: $\sqrt{ab} + \sqrt{bc} + \sqrt{ac} \leq a+b+c = 2p $ , but $ \frac{2p}{\sqrt{4p\cdot r}} = \sqrt{\frac{p}{r}} $

Ok I am using $s$ as semiperimeter because $p$ is commonly used for perimeter and that confuses me . Our desired is $$\frac1{\sqrt {2\sin A}} + \frac1{\sqrt {2\sin B}} + \frac1{\sqrt {2\sin C}}\leq\sqrt {\frac {s}{r}}$$Recall that Extended Law of Sines tells us that $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$which means that $$\frac{1}{2\sin A}=\frac{R}{a}$$and similarly. So it follows that our desired inequality is $$\sum_{cyc} \sqrt{\frac{R}{a}}\le \sqrt{\frac{s}{r}}$$so $$\sqrt{R}(\sum_{cyc}\frac{1}{\sqrt{a}})\le\sqrt{\frac{s}{r}}$$which tells us $$\sum_{cyc} \frac{1}{\sqrt{a}}\le\sqrt{\frac{s}{Rr}}$$We want to relate the circumradius, inradius, semiperiemter and side lengths...aha! We can use the fact that $$sr=[ABC]=\frac{abc}{4R}$$(this equality statement is easy to prove, I can prove it if anyone asks for it). Thus $Rr=\frac{abc}{4s}$, we can plug this into our inequality we just got to get $$\sum_{cyc} \frac{1}{\sqrt{a}}\le\sqrt{\frac{s}{Rr}}=\frac{2s}{\sqrt{abc}}$$So it suffices to show that $$\sum_{cyc} \sqrt{ab}\le 2s$$We know that $s=\frac{a+b+c}{2}$ so we can substitute this in to get that our desired is now $$\sum_{cyc} \sqrt{ab}\le a+b+c$$and this is true by trivial inequality after rearranging to $\frac{(\sqrt{a}-\sqrt{b})^2+(\sqrt{a}-\sqrt{c})^2+(\sqrt{c}-\sqrt{b})^2}{2}\ge0$ so we are done.