Solution:

Lemma 1. The area $S$ of a convex $n$-gon $A_{1}A_{2}\ldots A_{n}$ satisfies
$$2S \leq \frac{1}{2} \sum_{i=1}^{n} A_{i-1}A_{i+1} \cdot R_{i} = \sum_{i=1}^{n} R_{i}^{2} \sin A_{i}$$
where $R_{i}$ is the radius of the circumcircle of $\triangle A_{i-1}A_{i}A_{i+1}$, and the indices are reduced modulo $n$ (thus $A_{0} \equiv A_{n}$ and $A_{n+1} \equiv A_{1}$).

Proof. Let $M_{i}$ be the midpoint of $A_{i}A_{i+1}$, for $1 \leq i \leq n$. For each $i$ consider the quadrilateral formed by the segments $A_{i}M_{i}$ and $A_{i}M_{i-1}$, and by the perpendiculars onto these segments at $M_{i}$ and $M_{i-1}$, respectively.
We shall prove that these $n$ quadrilaterals cover our $n$-gon. Indeed, let $P$ be a point inside the $n$-gon. Let $PA_{k}$ be the least of the distances $PA_{i}$, $1 \leq i \leq n$. Having $PA_{k} \leq PA_{k+1}$ and $PA_{k} \leq PA_{k-1}$ means $P$ lies inside the $n$-gon and in each of the two half-planes containing $A_{k}$ and limited by the perpendicular bisectors of $A_{k}A_{k+1}$ and $A_{k}A_{k-1}$, thus in the $k$-th quadrilateral. To complete the proof, it remains to notice that the area of the $i$-th quadrilateral does not exceed $\frac{A_{i-1}A_{i+1}}{2} \cdot \frac{R_{i}}{2}$, and also that of course $\frac{A_{i-1}A_{i+1}}{2} = R_{i} \sin A_{i}$.

Lemma 2. If $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{5}$ are the angles of a convex pentagon, then $\sin^{2} \alpha_{1} + \sin^{2} \alpha_{2} + \cdots + \sin^{2} \alpha_{5} \leq 5 \sin^{2} 108^{\circ}$.

Proof. Will be presented at the end.

In the context of our problem, it follows that $2S \leq \sum_{i=1}^{5} R_{i}^{2} \sin A_{i}$. Using the Cauchy-Schwarz-Bunyakovsky inequality and Lemma 2 we now obtain
$$2S \leq \sum_{i=1}^{5} R_{i}^{2} \sin A_{i} \leq \sqrt{\left(\sum_{i=1}^{5} R_{i}^{4}\right)\left(\sum_{i=1}^{5} \sin^{2} A_{i}\right)} \leq \sqrt{5 \sin^{2} 108^{\circ} \sum_{i=1}^{5} R_{i}^{4}}$$
whence $\frac{4S^{2}}{5 \sin^{2} 108^{\circ}} \leq \sum_{i=1}^{5} R_{i}^{4}$.

Proof of Lemma 2. We may assume $0 < \alpha_{1} \leq \alpha_{2} \leq \cdots \leq \alpha_{5} < 180^{\circ}$. By a well-known formula $\alpha_{1} + \alpha_{2} + \cdots + \alpha_{5} = 540^{\circ}$. If $\alpha_{1} = 108^{\circ}$, then $\alpha_{2} = \cdots = \alpha_{5} = 108^{\circ}$, and the inequality becomes equality. If $\alpha_{1} < 108^{\circ}$, then $\alpha_{5} > 108^{\circ}$. Notice that $\alpha_{1} + \alpha_{5} < 270^{\circ}$ (if $\alpha_{1} + \alpha_{5} \geq 270^{\circ}$, then $\alpha_{2} + \alpha_{3} + \alpha_{4} \leq 270^{\circ}$, so $\alpha_{2} \leq 90^{\circ}$, a fortiori $\alpha_{1} \leq 90^{\circ}$, and therefore $\alpha_{5} \geq 180^{\circ}$, a contradiction).

Therefore we arrive at
$$\sin^{2} 108^{\circ} + \sin^{2}(\alpha_{1} + \alpha_{5} - 108^{\circ}) - \sin^{2} \alpha_{1} - \sin^{2} \alpha_{5} = 2 \cos(\alpha_{1} + \alpha_{5}) \sin(\alpha_{1} - 108^{\circ}) \sin(\alpha_{5} - 108^{\circ}) > 0.$$
This means that by the replacement of $\alpha_{1}$ with $108^{\circ}$ and of $\alpha_{5}$ with $\alpha_{1} + \alpha_{5} - 108^{\circ}$ we increase the sum of the squared sinuses. Repeating this operation, we will make all the angles equal to $108^{\circ}$, and the inequality is proved.