Best Embedding Constant for Weighted $L^p$ Space Let $\Omega \subset \mathbb{R}^n$ be a bounded Lebesgue measurable set of positive measure. Fix a function $A: \Omega \to \mathbb{R}$ such that $A \in L^{\infty}(\Omega)$ and there is a constant $a$ such that $A(x)>a>0$ for almost every $x \in \Omega$. Fix also a $p \in [1,\infty)$. We can norm the space $L^p(\Omega)$ in the following two ways: for every $f: \Omega \to \mathbb{R}$ , we set :
\begin{equation*}
\|f\| = (\int_{\Omega}|f(x)|^pdx)^{\frac{1}{p}}; \|f\|_A = (\int_{\Omega}|f(x)|^pA(x)dx)^{\frac{1}{p}}
\end{equation*}
Clearly $\|kf\|_A=|k|\|f\|_A$ for every $k \in \mathbb{R}$. By viewing $A(x)dx$ as a measure, Minkowski's inequality shows that $\|f+g\|_A\leq\|f\|_A+\|g\|_A$. Also, if $\|f\|_A=0$, then by the condition $A(x)>a>0$ almost everywhere we get $f=0$ almost everywhere. Therefore, $\|\cdot\|_A$ is a norm. Clearly, one always has:
\begin{equation*}
\|f\|_A \leq (\|A\|_{\infty})^{\frac{1}{p}} \|f\|
\end{equation*}
By assumption, $\|A\|_{\infty} \in (0,\infty)$. Therefore, $\| \cdot \|$ and $\| \cdot \|_A$ are equivalent norms in the Banach space $L^p(\Omega)$. The following best embedding constants are well-defined and finite:
\begin{equation*}
c = \inf \frac{\|f\|_A}{\|f\|}; d = \inf \frac{\|f\|}{\|f\|_A}
\end{equation*}
The infinums are taken over every nonzero $f\in L^p(\Omega)$.
My goal is to compute $c$ and $d$ by expressing them in terms of $\Omega$ and $A$ (or even better to find an $f$ attaining these infinums but this is not required). So let us just focus on computing $c$. Set $f(x)=1$ for every $x \in \Omega$ we see that:
\begin{equation*}
c \leqslant (\frac{1}{m(\Omega)}\int_{\Omega}A(x)dx)^{\frac{1}{p}}
\end{equation*}
My conjecture is that $c=(\frac{1}{m(\Omega)}\int_{\Omega}A(x)dx)^{\frac{1}{p}}$. But I cannot play it out using Hölder's inequalities or find counterexamples. How do I proceed?
Note1: as pointed out by Ryszard Szwarc, $A>0$ is not sufficient to make sure norms are equivalent:
let $\Omega=(0,1)$ and $A(x)=1-x.$ For $f_n(x)=x^{n/2}$ we have
$$\|f_n\|^2={1\over n+1},\qquad \|f_n\|^2_A={1\over (n+1)(n+2)}$$
In general let $$ U_n=\{x\in \Omega\,:\, |A(x)|\le n^{-1/2}\},\qquad m(U_n)>0$$. Then for $f_n=\chi_{U_n}$ we have
$$\|f_n\|^2=m(U_n), \qquad \|f_n\|^2_A\le {m(U_n)\over n}$$
I have updated the question to add the assumption that $A$ is bounded away from $0$.
Note2: if it is ever needed we can work in stronger set of hypothesis: $p \in (1,\infty)$, $\Omega$ open connected with smooth boundary or even $\Omega$ is a unit ball. But I guess these conditions do not really matter.
 A: The norms are not equivalent if $|A(x)|$ is not bounded away from $0.$
Indeed, let $p=2$, $\Omega=(0,1)$ and $A(x)=1-x.$ For $f_n(x)=x^{n/2}$ we have
$$\|f_n\|^2={1\over n+1},\qquad \|f_n\|^2_A={1\over (n+1)(n+2)}$$
In general let $$ U_n=\{x\in \Omega\,:\, |A(x)|\le n^{-1/2}\},\qquad m(U_n)>0$$ where $m$ denotes the Lebesgue measure. Then for $f_n=\chi_{U_n}$ we have
$$\|f_n\|^2=m(U_n), \qquad \|f_n\|^2_A\le {m(U_n)\over n}$$
Assume $A$ is bounded away from $0.$ Then $A^{-1}\in L^\infty.$ We have $$\|f\|_A\le \|A\|_\infty^{1/p}\|f\|_p,\qquad \|f\|_p\le \|A^{-1}\|_\infty^{1/p}\|f\|_A$$ The first inequality was shown by OP. The second inequality can be derived as follows:
$$\displaylines{\|f\|^p_p=\int\limits_\Omega |f(x)|^p\,dx=
\int\limits_\Omega |f(x)|^pA(x)A(x)^{-1}\,dx \\ \le \|A^{-1}\|_\infty  \int\limits_\Omega |f(x)|^pA(x)\,dx =\|A^{-1}\|_\infty\|f\|_A}$$ We obtain  $$c\ge |A^{-1}\|_\infty^{-1/p}\qquad d \ge |A\|_\infty^{-1/p}$$ For $N$ let $$\displaylines{U_n=\{x\in \Omega:\, |A(x)|\ge \|A\|_\infty (1-n^{-1})\} \\ V_n=\{x\in \Omega:\, |A^{-1}(x)|\ge \|A^{-1}\|_\infty(1- n^{-1})\}}$$
For $f=\chi_{U_n}$ and $g=\chi_{V_n}$ we get $$  {\|f\|_p\over \|f\|_A}\le\|A\|^{-1/p}_\infty (1-n^{-1})^{-1/p},\qquad {\|g\|_A\over \|g\|_p}\le\|A^{-1}\|^{-1/p}_\infty (1-n^{-1})^{-1/p}$$ Therefore  $$c= \|A^{-1}\|_\infty^{-1/p}\qquad d = \|A\|_\infty^{-1/p}$$
