Let $(E,\mathcal E,\lambda)$ be a $\sigma$-finite measure space, $f:E\to[0,\infty)^3$ be $\mathcal E$-measurable with $$\int|f|^2\:{\rm d}\lambda<\infty,\tag1$$ $k\in\mathbb N$ and $B_1,\ldots,B_k\in\mathcal E$ be disjoint with $$B:=\{f\ne0\}=\bigcup_{i=1}^kB_i\tag2$$ and $\lambda(B)\in(0,\infty)$.
Are we able to derive a bound $$\max_{1\le i\le k}\int_{B_i}\left|f-\frac1{\lambda(B_i)}\int_{B_i}f\:{\rm d}\lambda\right|^2\:{\rm d}\lambda\le c\int_B\left|f-\frac1{\lambda(B)}\int f\:{\rm d}\lambda\right|^2\:{\rm d}\lambda\tag3$$ for some $c\ge0$?
Using the idea described in the comments below this answer, it should be useful to note that $(3)$ can be rewritten as $$\max_{1\le i\le k}\lambda(B_i)\operatorname{Var}_{\operatorname P_i}[f]\le c\lambda(B)\operatorname{Var}_{\operatorname P}[f]\tag4,$$ where $$\operatorname P_i:=\frac{1_{B_i}\lambda}{\lambda(B_i)}$$ and $$\operatorname P:=\frac{1_B\lambda}{\lambda(B)}.$$
It is easy to check that $\int_\Omega |f - \sigma|^2 \, d \mu \leq \int \big|f - \frac{1}{\mu(\Omega)} \int_\Omega f\big|^2 \, d \mu$ for all $\sigma \in \Bbb{R}$.
Therefore, setting $\sigma_i := \frac{1}{\lambda(B_i)} \int_{B_i} f d \lambda$ and $\sigma := \frac{1}{\lambda(B)}\int_B f \, d \lambda$, we see $$ \int_{B_i} |f - \sigma_i|^2 d \lambda \leq \int_B |f - \sigma_i|^2 d \lambda \leq \int_B |f - \sigma|^2 \, d \lambda, $$ which proves your inequality with $c = 1$.