Let's denote by $A_{i}$ the point of tangency between circles $(O_i)$ and $(O_{i+1})$ and let $\theta_i=\angle O_iOA_i$. These angles satisfy $\theta_i \in \left(0,\frac{\pi}{2}\right)$ and $\theta_1+\cdots+\theta_n=\pi$. We can express $$r_i=\frac{R\sin \theta_i}{1-\sin \theta_i}$$ So both your inequalities can be rewritten in the form $$\frac{f(\theta_1)+\cdots+f(\theta_n)}{n}\geq f\left(\frac{\theta_1+\cdots+\theta_n}{n}\right)$$ with $f(x)=\frac{\sin x}{1-\sin x}$ and $f(x)=\left(\frac{\sin x}{1-\sin x}\right)^2$ respectively. Some calculus shows that both of these functions are convex in the desired interval so everything follows from Jensen's inequality.

Draw the tangents $OP_i$ and $OQ_i$ from $O$ to the circle $(O_i)$ and let $\theta_i=\frac{1}{2}\angle P_iOQ_i$. These angles satisfy $\theta_i \in \left(0,\frac{\pi}{2}\right)$ and $\theta_1+\cdots+\theta_n\geq\pi$. This is because the sectors $P_iOQ_i$ cover the whole plane. We can express $$r_i=\frac{R\sin \theta_i}{1-\sin \theta_i}$$ So both your inequalities can be rewritten in the form $$\frac{f(\theta_1)+\cdots+f(\theta_n)}{n}\geq f\left(\frac{\pi}{n}\right).$$ with $f(x)=\frac{\sin x}{1-\sin x}$ and $f(x)=\left(\frac{\sin x}{1-\sin x}\right)^2$ respectively. To prove this we can first establish $$\frac{f(\theta_1)+\cdots+f(\theta_n)}{n}\geq f\left(\frac{\theta_1+\cdots+\theta_n}{n}\right)$$ since both of these functions are convex in the desired interval so it follows from Jensen's inequality. And then the inequality $$f\left(\frac{\theta_1+\cdots+\theta_n}{n}\right)\geq f\left(\frac{\pi}{n}\right)$$ follows fromthe fact that both functions are increasing in this interval.