Here, for the weighted version of the Hardy Inequality, I refer to Muckenhoupt's formulation in Theorem 1 of 1

Sobolev Inequality: $$C_d \int_{\mathbb{R}^d} \vert \nabla \phi \vert^2 \geq \left( \int_{\mathbb{R}^d} \vert \phi(x)\vert^{2d/(d-2)} \right)^{\frac{d-2}{d}} \qquad $$