I think the following function $\Phi \colon S_{n,n} \times S_{n,n+2} \times S_{1,n+2} \to S_{n,n+1} \times S_{n,n+1} \times S_{1,n+1}$ is a surjection, which would prove the desired inequality. Each element of the domain and the range of $\Phi$ is a function on $\{1,\dots,n\} \cup \{1,\dots,n\} \cup \{1\}$; for the sake of our sanity let's write this as $\{1_a,\dots,n_a\} \cup \{1_b,\dots,n_b\} \cup \{1_c\}$.
Given $f\in S_{n,n} \times S_{n,n+2} \times S_{1,n+2}$, we define $\Phi(f) \in S_{n,n+1} \times S_{n,n+1} \times S_{1,n+1}$ as follows. (Hereafter, $j$ always denotes an integer between $1$ and $n$.)
- If $f(1_c) \ne n+2$, then set:
- $\Phi(f)(j_a) = \begin{cases} f(j_a), & \text{if } f(j_b) \ne n+2, \\ n+1, & \text{if } f(j_b) = n+2. \end{cases}$
- $\Phi(f)(j_b) = \begin{cases} f(j_b), & \text{if } f(j_b) \ne n+2, \\ f(j_a), & \text{if } f(j_b) = n+2. \end{cases}$
- $\Phi(f)(1_c) = f(1_c)$.
- If $f(1_c) = n+2$ and there exists $1\le j\le n$ such that $f(j_b) \ge n+1$, then let $k$ be the least such $j$; then set:
- $\Phi(f)(j_a) = \begin{cases} f(j_a), & \text{if } f(j_b) \le n, \\ n+1, & \text{if } f(j_b) \ge n+1. \end{cases}$
- $\Phi(f)(j_b) = \begin{cases} f(j_b), & \text{if } f(j_b) \le n, \\ n+1, & \text{if } f(j_b) \ge n+1. \end{cases}$
- $\Phi(f)(1_c) = \begin{cases} f(k_a), & \text{if } f(k_b) = n+1, \\ f(k_a)+1, & \text{if } f(k_b) = n+2. \end{cases}$
- If $f(1_c) = n+2$ but there does not exist $1\le j\le n$ such that $f(j_b) \ge n+1$, then I don't care what $\Phi(f)$ is.
The first part of the construction covers every function $g \in S_{n,n+1} \times S_{n,n+1} \times S_{1,n+1}$ except those for which there exists $j$ such that $g(j_a) = g(j_b) = n+1$; the second part (hopefully) covers these special functions $g$.