Yes, we claim that the inequality holds true. Let $A_1A_2...A_n$ be the given regular polygon, $B_1B_2...B_n$ be the given convex polygon and $O$ be the center of $\omega$. Consider $OM\bot A_1A_2$, then we have that $\left| A_1A_2 \right|=2\left| A_1 M\right|$ and $m(\widehat{A_1OM})=\frac{\pi}{n}$. Since $\left| A_1O \right|=R$, we get $\left| A_1 M\right|=R\sin\frac{\pi}{n}\Rightarrow \left| A_1A_2 \right|=2R\sin\frac{\pi}{n}$ and therefore $s=2nR\sin\frac{\pi}{n}$. We are asked to prove that $S\ge \frac{2sRn}{\sqrt{4n^2R^2-s^2}}=\frac{2sRn}{\sqrt{4n^2R^2\left( 1-\sin^2\frac{\pi}{n} \right)}}=\frac{s}{\cos\frac{\pi}{n}}=2Rn\tan\frac{\pi}{n}$. Let $\omega$ be tangent to $B_1B_2$ at the point $C_1$ and label $m(\widehat{B_1OC_1})=\alpha_1$, $m(\widehat{C_1OB_2})=\alpha_2$. Then we have $OC_1\bot B_1B_2$, $\left| OC_1 \right|=R$ and $\alpha_1,\alpha_2<\frac{\pi}{2}$. It's trivial in that in triangles $B_1OC_1$ and $C_1OB_2$, we get $\left| B_1C_1 \right|=R\tan\alpha_1$ and $\left| C_1B_2 \right|=R\tan\alpha_2$. Hence we have $\left| B_1B_2 \right|=R(\tan\alpha_1+\tan\alpha_2)$. With similar labeling, we get that $S=R\sum_{k=1}^{2n}\tan\alpha_k$ with $\sum_{k=1}^{2n}\alpha_k=2\pi$ and $\alpha_k<\frac{\pi}{2}$. Now the desired inequality is equivalent to $\frac{\sum_{k=1}^{2n}\tan\alpha_k}{2n}\ge\tan\frac{\pi}{n}$. Let $f(x)=\tan x$. Then $f''(x)=2\tan x\sec^2x\ge 0$ for $x<\frac{\pi}{2}$ and $f$ is convex. By Jensen's inequality we obtain $\frac{\sum_{k=1}^{2n}\tan\alpha_k}{2n}=\frac{\sum_{k=1}^{2n}f(\alpha_k)}{2n}\ge f\left( \frac{\sum_{k=1}^{2n}\alpha_k}{2n} \right)=\tan\frac{\pi}{n}$. Thus completing the proof.