I try to upper-bound an error term $X$ with two random variables $A$ and $B$, but I only have closed form for $\mathbf{E}[A] - \mathbf{E}[B] = c$, and I derived $X$ into the following term: $$ \mathbf{E}[X] \leq n - \dfrac{1}{1+\mathbf{E}\left[ \left(\dfrac{B}{A} \right)^2 \right]} $$ where $n$ is a constant.
I wonder if there are any tricks or inequalities to use to push this further?
So far I know that $A>0$, $B>0$, and $\mathbf{E}[A]>\mathbf{E}[B]$
Edit:
Thanks for Robert pointing out that the information is not enough, so let me add in all the information I have:
Given $s$ probabilities $\pi_1, \pi_2, \dots \pi_s$, which are sampled from the distribution $Beta(\alpha_s, \beta_s)$ respectively. And both $\alpha_s$ and $\beta_s$ are sampled from the same distribution $\mathcal{P}$ for all $s$.
For each probability $\pi_t$, we sample $n$ random variables with Binomial: $$ Y_{i,t} = Bin(2, \pi_t) $$ Therefore, there are total $ns$ Binomial random variables.
We define a set $\mathcal{T}$, whose element is a pair of indices. A pair of indices is in $\mathcal{T}$ if and only if the corresponding Binomial random variables are sampled from the same probability. In other words, for any two $Y_{i, t}$ and $Y_{j, t'}$, $(i, j) \in \mathcal{T}$ if and only if $t=t'$.
Now, back into the original question: $$ A = \dfrac{1}{m}\sum_{(i,j)\in\mathcal{T}}Y_{i,\cdot}Y_{j,\cdot} $$ $$ B = \dfrac{1}{m'}\sum_{(i,j)\not\in\mathcal{T}}Y_{i,\cdot}Y_{j,\cdot} $$ where $m$ is the number of elements in $\mathcal{T}$ and $m'$ is the total number of the rest pairs.
The given conditions on $A$ and $B$ don't let you say much about $E[(B/A)^2]$, only that it is positive.
Scenario 1: with probability $1$, $B = \varepsilon$ and $A = c + \varepsilon$, where $\varepsilon > 0$ is small. Then $E[(B/A)^2] = (\varepsilon/(c+\varepsilon))^2$ is small.
Scenario 2: $B=1$; with probability $1-\varepsilon$, $A = \varepsilon$ and with probability $\varepsilon$, $A = (1+c)/\varepsilon - 1 + \varepsilon$. The $E[(B/A)^2] \approx (1+2c)/\varepsilon^2$ is large.