The principal aim of this paper is to employ Bessel-type operators in proving the inequality begin{align*} int_0^pi dx , |f(x)|^2 geq dfrac{1}{4}int_0^pi dx , dfrac{|f(x)|^2}{sin^2 (x)}+dfrac{1}{4}int_0^pi dx , |f(x)|^2,quad fin H_0^1 ((0,pi)), end{align*} where both constants $1/4$ appearing in the above inequality are optimal. In addition, this inequality is strict in the sense that equality holds if and only if $f equiv 0$. This inequality is derived with the help of the exactly solvable, strongly singular, Dirichlet-type Schr{o}dinger operator associated with the differential expression begin{align*} tau_s=-dfrac{d^2}{dx^2}+dfrac{s^2-(1/4)}{sin^2 (x)}, quad s in [0,infty), ; x in (0,pi). end{align*} The new inequality represents a refinement of Hardys classical inequality begin{align*} int_0^pi dx , |f(x)|^2 geq dfrac{1}{4}int_0^pi dx , dfrac{|f(x)|^2}{x^2}, quad fin H_0^1 ((0,pi)), end{align*} it also improves upon one of its well-known extensions in the form begin{align*} int_0^pi dx , |f(x)|^2 geq dfrac{1}{4}int_0^pi dx , dfrac{|f(x)|^2}{d_{(0,pi)}(x)^2}, quad fin H_0^1 ((0,pi)), end{align*} where $d_{(0,pi)}(x)$ represents the distance from $x in (0,pi)$ to the boundary ${0,pi}$ of $(0,pi)$.
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