By my computation, I pose a conjecture as follows and I am looking for a proof:

Conjecture: Let $(O)$ be a circle with radius $R$, and $n$ be positive integer $n\ge 3$. Construct $n$ circles $(O_1), (O_2), \cdots, (O_n)$ external tangent to $(O)$ and $(O_i)$ tangent to $(O_{i+1})$ for $i=1$, $2$, $\cdots$, $n$ and $(O_{n+1}) \equiv (O_1)$. If radius of $(O_1)$, $(O_2)$, $\cdots$, $(O_n)$ are $r_1$, $r_2$, $\cdots$, $r_n$ respectively then

$$r_1+r_2+\cdots+r_n \ge \frac{n\sin^{\frac{\pi}{n}}}{1-\sin{\frac{\pi}{n}}} R$$


$$r_1^2+r_2^2+\cdots+r_n^2 \ge n\left( \frac{\sin{\frac{\pi}{n}}}{1-\sin{\frac{\pi}{n}}} \right)^2R^2$$

See also:

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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.

  • $\begingroup$ Can You see my comment in figure: i.stack.imgur.com/P6nFU.png ?? $\endgroup$ – Đào Thanh Oai Jul 20 '18 at 3:46
  • $\begingroup$ @ĐàoThanhOai thank you for catching the mistake. It should be fixed now. $\endgroup$ – Gjergji Zaimi Jul 20 '18 at 4:19
  • $\begingroup$ $f(x)=\left(\frac{\sin x}{1-\sin x}\right)^n$ is convex increasing in $(0. \frac{\pi}{2})$ with any $n \ge 1$. How can prove that? $\endgroup$ – Đào Thanh Oai Jul 20 '18 at 7:26

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