Let $$a, b, c$$ are the sides of a triangle

And $$\quad a+b=x, a b =y$$

Given, $$\quad x^2-c^2 =y$$

$$\begin{array}{lrl} \Rightarrow & (x+c)(x-c) & =y \\ \Rightarrow & (a+b+c)(a+b-c) & =y \\ \Rightarrow & 2 s(2 s-2 c) & =y \\ \Rightarrow & s(s-c) & =\frac{y}{4} \quad \text{... (i)}\\ \Rightarrow & \frac{s(s-c)}{y} & =\frac{1}{4} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{s(s-c)}{a b}=\frac{1}{4} \\ \Rightarrow & \cos ^2 \frac{C}{2}=\frac{1}{4} \\ \Rightarrow & \cos \frac{C}{2}= \pm \frac{1}{2} \\ \Rightarrow & \frac{C}{2}=\frac{\pi}{3}, \frac{2 \pi}{3} \\ \Rightarrow & \mathrm{C}=\frac{2 \pi}{3}, \frac{4 \pi}{3} \end{array}$$

$$\because \quad C=\frac{4 \pi}{3}$$ is impossible in a triangle

$$\therefore \quad C=\frac{2 \pi}{3}$$

Now, $$\frac{r}{\mathrm{R}}=\frac{(s-c) \tan \frac{\mathrm{C}}{2}}{\frac{c}{2 \sin \mathrm{C}}}$$

$$\begin{aligned} \Rightarrow \quad \frac{r}{\mathrm{R}} & =\frac{(s-c) \cdot \tan \frac{\pi}{3}}{\frac{c}{2 \sin \frac{2 \pi}{3}}} \\ \Rightarrow \quad \frac{r}{\mathrm{R}} & =\frac{\frac{y}{4 s} \cdot \sqrt{3}}{\frac{c}{2 \cdot \frac{\sqrt{3}}{2}}} \quad \left[(s-c)=\frac{y}{4 s} \text { from equation (i) }\right] \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad \frac{r}{\mathrm{R}}=\frac{3 y}{2 \cdot 2 s \cdot c} \\ & \Rightarrow \quad \frac{r}{\mathrm{R}}=\frac{3 y}{2(a+b+c) c} \\ & \Rightarrow \quad \frac{r}{\mathrm{R}}=\frac{3 y}{2(x+c) \cdot c} \end{aligned}$$

Hint:

$$\Rightarrow$$ In a triangle $$\triangle \mathrm{ABC}$$

(i) $$r=(\mathrm{s}-c) \tan \frac{\mathrm{C}}{2}, \mathrm{R}=\frac{c}{2 \sin \mathrm{C}}$$

(ii) $$\cos \frac{c}{2}=\sqrt{\frac{s(s-c)}{a b}}, s=\frac{a+b+c}{2}$$