I was given the question following question:
Suppose $A$ and $B$ are $n\times n$ matrices with entries in $\mathbb R$. Show that $\lVert AB\rVert_{HS} \le \lVert A\rVert_{HS}\lVert B\rVert_{HS}.$ [Hint: you may need Cauchy-Schwarz]
Given the definition of the Hilbert-Schmidt inner product and the hint, I was able to arrive at $\lVert AB\rVert_{HS}^2 \le \lVert A^{\top}A\rVert_{HS}\lVert BB^{\top} \rVert_{HS}$ which narrows the problem down to showing $\lVert A^{\top}A\rVert_{HS} \le \lVert A\rVert_{HS}^2$ (the case for $BB^{\top}$ is similar). Given that $A^{\top}A$ is symmetric, this can be shown if I can prove that all the real eigenvalues of that matrix are non-negative, however I'm unsure if this is true, or how to approach that proof if it is true.
For context, I'm not in a functional analysis class, so I would expect the intended solution to use relatively basic results from linear algebra.