I have a question about the following problem:
Suppose
\begin{equation}\label{one}\tag{1} \sqrt{n}\left(\begin{bmatrix}X_n\\Y_n\\Z_n\end{bmatrix} - \begin{bmatrix}\mu_x\\\mu_y\\\mu_z\end{bmatrix}\right) \xrightarrow[]{D} \mathcal{N}\left(\begin{bmatrix}0\\0\\0\end{bmatrix} - \begin{bmatrix} \sigma^2_{x} & \sigma_{xy} & \sigma_{xz} \\ \sigma_{xy} & \sigma^2_{y} & \sigma_{yz} \\ \sigma_{xz} & \sigma_{yz} & \sigma^2_{z} \end{bmatrix}\right) \end{equation}
Show using Slutksy's Theorem that
\begin{equation} \sqrt{n}(X_n + Y_nZ_n - \mu_x - \mu_y\mu_z) \xrightarrow[]{D} \mathcal{N}(0,\sigma^2) \end{equation}
for some $\sigma^2$, and find $\sigma^2$.
I am quite certain this can be solved using the Delta method. However, the problem instructs to make use of Slutsky's Theorem. Now in our textbook "A Course on Large Sample Theory" by T.S. Ferguson, the Delta method is proved using Slutsky's Theorem. Is there any other way to show this without simply copying that proof and then using the Delta method?
I was thinking that maybe if you show that the assumption implies that
\begin{align} \begin{split} X_n &\xrightarrow[]{D} \mu_x, \\ Y_n &\xrightarrow[]{D} \mu_y, \\ Z_n &\xrightarrow[]{D} \mu_z, \end{split}\label{two}\tag{2} \end{align}
then by Slutsky's Theorem, also $Y_nZ_n \xrightarrow[]{D} \mu_y\mu_z$ and therefore again by Slutsky's Theorem $X_n + Y_nZ_n \xrightarrow[]{D} \mu_x + \mu_y\mu_z$. Then, since $\mu_x + \mu_y\mu_z$ is constant, also
\begin{equation}\label{three}\tag{3} X_n + Y_nZ_n \xrightarrow[]{P} \mu_x + \mu_y\mu_z \end{equation}
Now, if I show that \eqref{one} implies \eqref{two}, and furthermore that \eqref{three} implies that $X_n+Y_nZ_n$ has a finite second (and hence first) moment the proof is complete. I sadly don't see how to do that. Any help?