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 <a href="https://en.wikipedia.org/wiki/Jensen%27s_inequality">Jensen's inequality</a>. 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.