Given a normalized real vector $\theta \in \mathbb{R}^n$ (its elements sum up to $1$, i.e., $||\theta||_1 = 1$), an integer vector $x \in \mathbb{Z}_{\geq 0}^n$ and an integer $S \geq \sum\limits_{i=1}^n x_i$, prove or disprove that
$$S! \sum\limits_{\sum y_i = S; x_i \leq y_i} \prod \limits_{i=1}^n\frac{\theta_i^{y_i}}{y_i !} \geq \prod\limits_{i=1}^n \sum\limits_{t=x_i}^{S - \sum\limits_{j=1}^{i-1}x_j}\binom{S - \sum\limits_{j=1}^{i-1}x_j}{t}\theta_i^t(1-\theta_i)^{S - \sum\limits_{j=1}^{i-1}x_j - t}$$
In other words, we need to estimate the probability of distributing $S$ balls in $n$ bins such that the $i$-th bin contains at least $x_i$ balls (LHS), via a product of binomial probabilities (RHS).
For example, $\theta = [0.3, 0.7]$, $x = [0, 1]$, $S = 3$.
$LHS = 1-(0.3)^3 = 0.973$.
$RHS = 1 \times (0.7)^0 (0.3)^3 =0.973$.
$x_1 = 0, S - \sum\limits_{j=1}^{0}x_j = 3$;
$x_2 = 1, S - \sum\limits_{j=1}^{1}x_j = 3$;
$LHS \geq RHS$.
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Numerical experiments suggest that the inequality is True. The RHS can be directly calculated, and the LHS can be calculated in $O(nSlogS)$ time via divide-and-conquer+FFT. However, I have no idea how to prove or disprove it.
Related MSE question: HERE.