1. The problem involves the relation between the inradius $r$, circumradius $R$, and sides $a,b,c$ of a triangle, given by the equation: $$2R = \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$$ 2. We are also given the formulas: - Circumradius: $$R = \frac{abc}{4\Delta}$$ - Inradius: $$r = \frac{\Delta}{s}$$ where $\Delta$ is the area of the triangle and $s$ is the semiperimeter. 3. The goal is to simplify and verify the given expression or find a relation involving $R$, $r$, $a$, $b$, $c$, and $\Delta$. 4. Start by rewriting the right-hand side (RHS): $$\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{c + a + b}{abc} = \frac{a + b + c}{abc}$$ 5. The left-hand side (LHS) is $2R$, so the equation becomes: $$2R = \frac{a + b + c}{abc}$$ 6. Substitute $R = \frac{abc}{4\Delta}$ into the LHS: $$2 \times \frac{abc}{4\Delta} = \frac{a + b + c}{abc}$$ which simplifies to: $$\frac{abc}{2\Delta} = \frac{a + b + c}{abc}$$ 7. Cross-multiplied: $$ (abc)^2 = 2\Delta (a + b + c) $$ 8. Using the semiperimeter $s = \frac{a + b + c}{2}$, rewrite: $$ (abc)^2 = 4\Delta s $$ 9. Recall the inradius formula $r = \frac{\Delta}{s}$, so $\Delta = r s$. 10. Substitute $\Delta$: $$ (abc)^2 = 4 r s^2 $$ This is a derived relation connecting the sides, area, inradius, and semiperimeter. Final answer: $$2R = \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a + b + c}{abc}$$ and $$R = \frac{abc}{4\Delta}, \quad r = \frac{\Delta}{s}$$ with the relation: $$(abc)^2 = 4 r s^2$$