I encountered the following which is claimed to be true, but I don't know how to prove it. Also, I wonder whether it holds in wider context.
I am now considering symmetric matrix $A,M\in\mathbb{R}^{n\times n}$. W.l.o.g. I can assume(with SVD) that $M=diag\{\sigma_1(M),\cdots,\sigma_k(M),0,\cdots,0\}$ with some $k<n$ and with $\sigma_i(M)$-s being the singular values of $M$ in non-increasing order. Also assume the singular values of $A$ in non-increasing order are $\sigma_1(A),\cdots,\sigma_n(A)$. Then I need to show
$$\langle A,M\rangle_{HS}\leq\sum_{i=1}^k\sigma_i(A)\sigma_i(M).$$
I tried using the SVD of $A$(assume $A^\top=U\Sigma V, U=(u_1,\cdots,u_n)^\top,V=(v_1,\cdots,v_n)$): \begin{align*} \langle A,M\rangle_{HS}=Tr(A^\top M) &=\sum_{i=1}^k\sigma_i(M)Tr(A^\top diag\{0,\cdots,0,1_i,0,\cdots,0\})\\ &= \sum_{i=1}^k\sigma_i(M)Tr(U\Sigma V diag\{0,\cdots,0,1_i,0,\cdots,0\})\\ &= \sum_{i=1}^k\sigma_i(M)Tr(\Sigma (V diag\{0,\cdots,0,1_i,0,\cdots,0\}U))\\ &= \sum_{i=1}^k\sigma_i(M)Tr(\Sigma v_iu_i^\top)\\ &= \sum_{i=1}^k\sigma_i(M)(u_i^\top\Sigma v_i), \end{align*}
but then I have no idea how to show the desired inequality.
Meanwhile, I wonder whether this inequality holds for any matrices $A,M$ with the same size.
