How to prove that the angle between two sides of that triangle is less than 60 degree? The product of two sides of triangle is equal to 8*(R*r) where R is circumradius of this triangle, and r is inradius of this triangle.
How to prove that the angle between two sides of that triangle is less than 60 degree? 

Suppose $ab = 8Rr$. We have $abc = 4RA$ where $A$ is the area of the triangle, hence $2rc = A$. The area of the triangle is also given by $rs$ where $2s=a+b+c$. It follows that $2rc = rs$ thus $4c=a+b+c$ or $3c=a+b$. It follows that $$ \cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{9a^2+9b^2-(a+b)^2}{18ab} = \frac{8a^2+8b^2-2ab}{18ab} \geq \frac{14}{18}, $$
(using the fact that $a^2+b^2 \geq 2ab$)
implying that $C \leq \arccos(\frac{14}{18}) = 38.9^\circ$.
