a geometric inequality of circumscribed polygon I learned the following theorem from Roger Johnson's book Advanced Euclidean Geometry §101g:

Theorem:  If a polygon is inscribed in a circle, and A is a point of
the circle, the product of the distances from A to the sides of the
polygon, equals the product of the distances from A to the tangents of
the circle at the vertices of the polygon.

I used the geometer's sketchpad and found that the following inequality may be true:
If the point A is inside the circle, the first product is always less than the second product, regardless of the convexity of the polygon.

The first product is less than the second product,iff A is within the region coloured purple

The same inequality is true for concave polygons
(Since my mother tongue is not English, if there is any unclear expression, just ask me!)
 A: Your inequality is true.
Let $\mathcal{C}$ be any circle of radius $r$. For any $P, Q \in \mathcal{C}$, let

*

*$\ell_P$ and $\ell_Q$ be the tangent lines of $\mathcal{C}$ at $P$ and $Q$,

*$\ell_{PQ}$ be the line passing  through $P$ and $Q$.

Choose a Cartesian coordinate system for the Euclidean plane so that $\mathcal{C}$ is centered at origin with $P, Q$ position symmetrically respect to the $x$-axis. i.e. for some suitably chosen $\theta \in (0,\pi)$, $P, Q$ are positioned at $(r\cos\theta, r\sin\theta)$ and $(r\cos\theta,-r\sin\theta)$.
In this coordinate system, the equations for the lines are
$$\begin{array}{rc}
\ell_P :& \cos\theta x + \sin\theta y - r = 0\\
\ell_Q :& \cos\theta x - \sin\theta y - r = 0\\
\ell_{PQ} :& x - r\cos\theta = 0
\end{array}$$
Let $\mathcal{D}$ be the open disk bounded by $\mathcal{C}$. If $A = (u,v) \in \mathcal{D}$, we will have $r^2 > u^2 + v^2$. The distances of $A$ to the lines will be given by the formulae:
$$\begin{align}
d(A,\ell_P) &= r - ( \cos\theta u + \sin\theta v )\\
d(A,\ell_Q) &= r - ( \cos\theta u - \sin\theta v )\\
d(A,\ell_{PQ}) &= | u -  r\cos\theta |
\end{align}$$
With a little bit of algebra, we find
$$\begin{align}
d(A,\ell_P)d(A,\ell_Q) - d(A,\ell_{PQ})^2
&=
(r - \cos\theta u)^2 - v^2\sin^2\theta - (u- r\cos\theta)^2\\
&= \sin^2\theta(r^2 - u^2-v^2)\\ 
&> 0\end{align}$$
As a result,
$$\sqrt{d(A,\ell_P)d(A,\ell_Q)} > d(A,\ell_{PQ})$$
Let $P_1,P_2,\ldots,P_n$ be any $n$ points on $\mathcal{C}$ and use $P_0$ as an alias of $P_n$. Substitute $(P,Q)$ by following $n$ pairs of points $(P_n,P_1), (P_1,P_2),\ldots,(P_{n-1},P_n)$, your inequality follows:
$$\prod_{k=1}^n d(A,\ell_{P_k})
= \prod_{k=1}^n \sqrt{d(A,\ell_{P_{k-1}})d(A,\ell_{P_k})}
> \prod_{k=1}^n d(A,\ell_{P_{k-1}P_k})$$
If you replace $\mathcal{D}$ by the closed disk $\bar{\mathcal{D}}$, the $>$ in above inequality becomes $\ge$.
