Inequality of Denominators

Multiply $a+b+c$ on both sides and simplify to get $6+\sum \dfrac{a+b}{c}+\sum\dfrac{a}{b+c}\ge k$

We see that $\sum \dfrac{a+b}{c} =\sum \dfrac{a}{c}+\dfrac{b}{c} \ge 6\sqrt[6]{\dfrac{a}{b}\cdot \dfrac{b}{a}\cdot\dfrac{a}{c}\cdot \dfrac{c}{a}\cdot\dfrac{c}{b}\cdot \dfrac{b}{c} }=6$ by six-variable AM-GM.

Also, $\sum\dfrac{a}{b+c}\ge \dfrac{3}{2}$ by Nesbitt's Inequality.

So $6+6+\dfrac{3}{2} \ge k\implies k\le \dfrac{27}{2}$ and our answer is $27+2=\boxed{29}$ .

By AM-HM $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \leq \frac{a+b+c}{3}$ and $\frac{3}{\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}} \leq \frac{2(a+b+c)}{3}$ .

So: $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 9$ and $(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c})\geq \frac{9}{2}$ .

Adding gives $k=\frac{27}{2}$ .

Equality is obtained iff $a=b=c$

Ok,the solutions here are so creative. Maybe mine one is very poor but if you substitute a=b=c=1 since the value of k is independent of it. After simplifying , we get k is less than or equal to 27/2 where p= 27, q= 2 which adds upto 29. LOL.Not creative but time saving.

Here's my solution:

We shall normalize by assuming $a+b+c=1$ . Then, the inequality reduces to $\sum_{cyc} \frac{1}{a}+\sum_{cyc} \frac{1}{1-a} \geq k$ now, let $f(x)=\frac{1}{x}+\frac{1}{1-x}$ . Since $f(x)$ a convex function, by Jensen's Inequality, we have $f(\frac{a+b+c}{3}) \leq \frac{f(a)+f(b)+f(c)}{3}$ \] so, $f(a)+f(b)+f(c) \geq \frac{27}{2}$ . But we also know that $\sum_{cyc} \frac{1}{a}+\sum_{cyc} \frac{1}{1-a}=f(a)+f(b)+f(c)$ . And hence we have $k=\frac{27}{2}$ , so $p+q=27+2=\boxed{29}$