Longest side is in interval

Without loss of generality, let $a \le b \le c$ , then $\max(a,b,c) =c$ . And let $b = a+\delta$ and $c = a+\delta+\epsilon$ , where $\delta, \epsilon \ge 0$ . Then let $p = \dfrac {\color{#3D99F6}c}{a+b+c} = \dfrac {\color{#3D99F6}a+\delta+\epsilon}{a+b+c}$ . Therefore, $p$ is minimum, when $\delta=\epsilon=0$ , then $a=b=c$ and $p = \dfrac {a}{a+b+c} = \dfrac a{3a} = \dfrac 13$ .

By triangle inequality , we have $a+b > c$ and the limit is reached when $\angle C \to 180^\circ$ , then $c \to a+b$ and $q = \dfrac {\color{#3D99F6}c}{a+b+c} = \dfrac {\color{#3D99F6}a+b}{a+b+{\color{#3D99F6}a+b}} = \dfrac 12$ .

Therefore, $p+q=\dfrac 13+\dfrac 12 = \dfrac 56$ ; $\implies x+y = 5+6 = \boxed{11}$ .