The minimum is irrational?

Notice that $\frac{1}{\sqrt{x}}$ has second derivative $\frac{3}{4 \sqrt{x^5}}$ (which is positive for all positive x) and is therefore convex. Since $a+b+c = 1$ , we can apply jensen's inequality, as follows:

$\large \dfrac{c}{\sqrt{ab + 1 - c}} + \dfrac{b}{\sqrt{ca + 1 - b}} + \dfrac{a}{\sqrt{bc + 1 - a}} \geq \frac{1}{\sqrt{3abc + a + b + c - a^2 - b^2 - c^2}}$

Now, notice that since $a+b+c = 1$ , we have $a+b+c = (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ac$ .

Now, substituting this into the expression gives us

$\frac{1}{\sqrt{3abc + a + b + c - a^2 - b^2 - c^2}} = \frac{1}{\sqrt{3abc +(a+b+c)^2 - a^2 - b^2 - c^2}} = \frac{1}{\sqrt{3abc+2ab+2bc+2ac}}$ .

Also, notice that, since $a+b+c = 1$ , we have $1 = (a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca \geq 3ab + 3bc + 3ca$ ,

Giving us $ab + bc + ca \leq \frac{1}{3}$ .

Also, $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$ , giving us $abc \leq \frac{1}{27}$

Piecing it all together, we have

$3abc + 2ab + 2bc + 2ca \leq \frac{3}{27} + \frac{2}{3} = \frac{7}{9}$ .

This implies that

$\frac{1}{\sqrt{3abc+2ab+2bc+2ac}} \geq \frac{1}{\sqrt{\frac{7}{9}}} = \frac{3\sqrt{7}}{7}$ .

Thus, we have the chain of inequalities

$\dfrac{c}{\sqrt{ab + 1 - c}} + \dfrac{b}{\sqrt{ca + 1 - b}} + \dfrac{a}{\sqrt{bc + 1 - a}} \geq \frac{1}{\sqrt{3abc + a + b + c - a^2 - b^2 - c^2}} = \frac{1}{\sqrt{3abc+2ab+2bc+2ac}} \geq \frac{1}{\frac{7}{9}} = \frac{3\sqrt{7}}{7}$ , with equality when $a = b = c$ .

Thus, our answer is $3+7+7 = \boxed{17}$ .

By Hölder's inequality, we have

$\displaystyle (\sum_{cyc} \frac{c}{\sqrt{ab+1-c}})(\sum_{cyc} \frac{c}{\sqrt{ab+1-c}})(\sum_{cyc} c(ab+1-c)) \geq [\sum_{cyc} (\frac{c}{\sqrt{ab+1-c}})(\frac{c}{\sqrt{ab+1-c}})(c(ab+1-c))^{\frac{1}{3}}]^{3}$

$= \displaystyle (\sum_{cyc} c)^{3} = 1$ . Also, we have

$\displaystyle \sum_{cyc} c(ab+1-c) = 3abc+(a+b+c) - (a^2 + b^2+c^2) \leq 3(\frac{a+b+c}{3})^{3} + 1 - \frac{(a+b+c)^2}{3} = \frac{1}{9} + 1 - \frac{1}{3} = \frac{7}{9}$ , due to AM-GM.

Therefore $\displaystyle \sum_{cyc} \frac{c}{\sqrt{ab+1-c}} \geq \frac{1}{\sqrt{\sum_{cyc} c(ab+1-c)}} = \frac{3}{\sqrt{7}} = \frac{3\sqrt{7}}{7}$ . If equality holds, then we need $a=b=c$ , because we applied AM-GM on $a,b,c$ . Indeed, at $a=b=c=\frac{1}{3}$ ,the expression takes the value of $\frac{3\sqrt{7}}{7}$ . So $\boxed{p=3 , q=7 , r=7}$ , and $p+q+r = \boxed{17}$ .