Let $(\Omega,\mathcal A)$ be a measurable space, $\mu$ be a signed measure on $(\Omega,\mathcal A)$, $\nu$ be a measure on $(E,\mathcal E)$ and $c\ge0$ with $$\left|\mu(E)\right|\le c\sqrt{\nu(E)}\;\;\;\text{for all }E\in\mathcal E\tag1$$ for some $\cap$-stable $\mathcal E\subseteq\mathcal A$ with $\sigma(\mathcal E)=\mathcal A$.
Now, let $|\mu|$ denote the variation of $|\mu|$, i.e. $$|\mu|(A):=\sup\left\{\sum_{i=1}^k\left|\mu(A_i)\right|:k\in\mathbb N\text{ and }A_1,\ldots,A_k\in\mathcal A\text{ are mutually disjoint with }\biguplus_{i=1}^kA_i\subseteq A\right\}$$ for $A\subseteq\Omega$.
Are we able to show $$|\mu|(A)\le c\sqrt{\nu(A)}\tag2$$ for all $A\in\mathcal A$?
Remark: I know how I can conclude $|\mu|(A)\le\sqrt{\nu_1(A)}\sqrt{\nu_2(A)}$ for all $A\in\mathcal A$ from $|\mu(E)|\le\sqrt{\nu_1(E)}\sqrt{\nu_2(E)}$ for all $E\in\mathcal E$ (where $\nu_i$ is a measure on $(E,\mathcal E)$), but in the case of $(1)$ and $(2)$ we cannot apply the Cauchy-Schwarz inequality.
This is not true, even when $\mu = \nu.$ For concreteness let $\mu = \nu = \mathcal L^n$ be the Lebesgue measure on $\mathbb R^n$ equipped with the Borel $\sigma$-algebra, and consider, $$ \mathcal{E} = \{ E \subset \mathbb R^n : Q \text{ Borel with } \mathcal L^n(Q) \leq 1 \}. $$ This family is certainly stable under finite unions, and for any $E \in \mathcal{E}$ we have, $$ \mathcal L^n(E) \leq \sqrt{\mathcal L^n(E)}, $$ since $\sqrt{t} - t \geq 0$ for $0 \leq t \leq 1.$ However this inequality does not hold for any $A \subset \mathbb R^n$ Borel with $1 < \mathcal L^n(A) < \infty,$ since $\sqrt{t} - t < 0$ whenever $t>1.$