3 typo fixed

The answer is no. Consider $B\subset A$ with $\mu(B)$ very small, $C_1=A\setminus B$ and $C_2=X\setminus C_1$. Then the r.h.s. equals $\mu(B)^2/\mu(A)\mu(C_2)$ which can be less than $\mu(B)$ since $\mu(A)$ and $\mu(C_2)$ can be grater than $1/2$ and $\mu(B)$ less than $1/10$.

Added later. Under the assumption that each $A$ an $B$ is a union of some $C_\sigma$, each $C_\sigma$ is contained in either $A\cap B$ or $A\setminus B$ or $B\setminus A$ or $X\setminus(A\cup B)$. Only those contained in $A\cap B$ contribute to the r.h.s., and hence the r.h.s. equals $$\sum_{\sigma:C_\sigma\subset A\cap B} \frac{\mu(C_\sigma)}{\mu(A)} = \frac{\mu(A\cap B)}{\mu(A)}\le B)}{\mu(A)}\ge \mu(A\cap B)$$ because $\mu(A)\le 1$.

The answer is no. Consider $B\subset A$ with $\mu(B)$ very small, $C_1=A\setminus B$ and $C_2=X\setminus C_1$. Then the r.h.s. equals $\mu(B)^2/\mu(A)\mu(C_2)$ which can be less than $\mu(B)$ since $\mu(A)$ and $\mu(C_2)$ can be grater than $1/2$ and $\mu(B)$ less than $1/10$.
Added later. Under the assumption that this strange inequality could possibly be true..each $A$ an $B$ is a union of some $C_\sigma$, each $C_\sigma$ is contained in either $A\cap B$ or $A\setminus B$ or $B\setminus A$ or $X\setminus(A\cup B)$. Only those contained in $A\cap B$ contribute to the r.h.s., and hence the r.h.s. equals $$\sum_{\sigma:C_\sigma\subset A\cap B} \frac{\mu(C_\sigma)}{\mu(A)} = \frac{\mu(A\cap B)}{\mu(A)}\le \mu(A\cap B)$$ because $\mu(A)\le 1$.
The answer is no. Consider $B\subset A$ with $\mu(B)$ very small, $C_1=A\setminus B$ and $C_2=X\setminus C_1$. Then the r.h.s. equals $\mu(B)^2/\mu(A)\mu(C_2)$ which can be less than $\mu(B)$ since $\mu(A)$ and $\mu(C_2)$ can be grater than $1/2$ and $\mu(B)$ less than $1/10$.