Tail bound on trace norm / nuclear norm / Schatten-1 norm of Rademacher matrix

Question

Let $0 < r \leq d$ integers. Let $X$, $Y$ be $d \times r$ matrices of independent Rademacher variables, that is, $X,Y \in \mathbb{R}^{d \times r}$ with entries $\pm1$ with probability $1/2$. I am wondering about the distribution of the trace norm / nuclear norm / Schatten-$1$ norm of $X+Y$, denoted by $\|X+Y\|_1 = \mathrm{Tr}[\sqrt{(X+Y)^{\dagger}(X+Y)}]$. In particular, I would like to know if there exist constants $c,C,K > 0$ such that $$\mathbb{P}\left[\|X+Y\|_1 \leq r\sqrt{d}/K\right] \leq C\exp(-crd).$$

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Here is my attempt so far:

1. The squared Frobenius norm is the sum of squares of the matrix entries, and so $\|X+Y\|_2^2$ is simply the sum of $dr$ variables which are independently $0$ or $4$ with probability $1/2$. Thus, by Hoeffding's inequality, $$\begin{align*}\mathbb{P}\left[\|X+Y\|_2^2 \leq dr\right] &= \mathbb{P}\left[\|X+Y\|_2^2 - 2dr\leq -dr\right] \\ &\leq \exp(-2(dr)^2/(16dr)) = \exp(-dr/8).\end{align*}$$
2. For the operator norm $\|X\|$, there exist positive constants $c,K > 0$ such that $\mathbb{P}[\|X\| \geq K\sqrt{d}] \leq 2\exp(-cd)$ (Proposition 2.4 of this paper). Similarly for $\|Y\|$.
3. Combining the above statements with the inequality $\|X+Y\|_2^2 \leq \|X+Y\|_1 \cdot \|X+Y\|$, we obtain through a union bound that $$\mathbb{P}\left[\|X+Y\|_1 \leq r\sqrt{d}/(2K)\right] \leq \exp(-dr/8) + 4\exp(-cd).$$

Hence, the question that remains is if the right-hand term in the tail bound can be strengthened to something that grows as $\exp(-crd)$, rather than $\exp(-cd)$.

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Numerical simulations seem to suggest that the result is not true if we replace $X + Y$ with just a single Rademacher matrix $X$.