$\dfrac{R}{r}$

Realize that the area of a triangle can be written in a few key ways, namely

$\begin{aligned} A & = rs &(1)\\ & = \frac{abc}{4R} &(2)\\ & = \sqrt{s(s-a)(s-b)(s-c)} &(3) \end{aligned}$

where $s$ is the semi perimeter of the triangle and $a = 4k$ , $b = 5k$ , $c = 6k$ where $k$ is some positive integer. Because $s = \frac{4k + 5k + 6k}{2} = \frac{15}{2}k$ , we can solve for the area using equation $(3)$ .

$\begin{aligned} A & = \sqrt{\Big(\frac{15}{2}k\Big)\Big(\frac{7}{2}k\Big)\Big(\frac{5}{2}k\Big)\Big(\frac{3}{2}k\Big)} \\ & = \sqrt{\frac{1575}{16}k^{4}} \\ & = \frac{15\sqrt{7}}{4}k^{2} \end{aligned}$

Now, we can solve for $r$ using equation $(1)$ as such $\begin{aligned} r & = \frac{\frac{15\sqrt{7}}{4}k^{2}}{\frac{15}{2}k}\\ & = \frac{\sqrt{7}}{2}k \end{aligned}$

We can also solve for $R$ using equation $(2)$ as such $\begin{aligned} R & = \frac{(4k)(5k)(6k)}{(4)(\frac{15\sqrt{7}}{4}k^{2})} \\ & = \frac{30}{\frac{15\sqrt{7}}{4}}k \\ & = \frac{8}{\sqrt{7}}k \\ & = \frac{8\sqrt{7}}{7}k \end{aligned}$

Upon finding $R$ and $r$ , we can compute $\frac{R}{r}$ as such

$\begin{aligned} \frac{R}{r} & = \frac{\frac{8\sqrt{7}}{7}k}{\frac{\sqrt{7}}{2}k} \\ & = \frac{16}{7} \end{aligned}$

Thus, $a + b = 16 + 7 = \boxed{23}$