This is, except for the finishing touches, not a pentagon inequality, just the special case of an $n$-gon inequality. I think it might have been more instructive to propose it without the extra Cauchy-Schwartz at the end. Let's start with a convex polygon $A_1A_2\dots A_n$ whose area is $S$. Define $A_{i\pm n}=A_i$ and let $O_i$ be the circumcenter of triangle $A_{i-1}A_iA_{i+1}$. Let $R_i=O_iA_i$ and let $\alpha_i=\angle A_{i-1}A_iA_{i+1}$. Consider the midpoints of the edges: let $M_i$ be the midpoint of $A_iA_{i-1}$. Note that $O_iM_i$ and $O_iM_{i+1}$ are the perpendicular bisectors of sides $A_iA_{i-1}$ and $A_iA_{i+1}$. Now look at the quadrilaterals $M_iA_iM_{i+1}O_i$ for $i=1,2,\dots,n$. They cover the entire $n$-gon, as any interior point $P$ is covered - WLOG if it is closest to vertex $A_1$, then $PA_1\le PA_2$ and $PA_1\le PA_3$, which is equivalent to $P$ being within $M_iA_iM_{i+1}O_i$, provided $P$ is within the $n$-gon. It follows that $$S\le \sum_{i=1}^n [M_iA_iM_{i+1}O_i].$$ However, if the distance of $O_i$ to $M_iM_{i+1}$ is $x$ and the distance of $A_i$ to $M_iM_{i+1}$ is $y$, then $[M_iA_iM_{i+1}O_i]=\frac{x+y}2 M_iM_{i+1}$, where $M_iM_{i+1}=\frac12 A_{i-1}A_{i+1}=\frac12\cdot 2R_i\sin \alpha_i$ and $x+y\le R_i$ implies $$2S\le \sum_{i=1}^n R_i^2\sin \alpha_i.$$ To derive the proposed inequality, it suffices to apply Cauchy's inequality: $$4S^2\le (\sum R_i^2\sin\alpha_i)^2\le (\sum R_i^4)(\sum \sin^2\alpha_i),$$where for $n=5$ we surely have $$\frac1n\sum \sin^2\alpha_i\le \sin^2(\frac1n\sum\alpha_i)=\sin^2 \frac{(n-2)\pi}n,$$which yields the result $$\frac{4S^2}{n\sin^2 \frac{(n-2)\pi}n}\le \sum_{i=1}^n R_i^4.$$ To prove that pesky Jensen-like inequality (note that $\sin^2 x=\frac{1-\cos 2x}2$ is not concave on $[0;\pi]$), just show that under certain circumstances, we may change the two angles $\alpha<\frac{(n-2)\pi}n<\beta$ to $\alpha+\beta-\frac{(n-2)\pi}n$ and $\frac{(n-2)\pi}n$ in a way that the LHS increases but the RHS is invariant. (I'll leave this to the reader.)