Singular-value inequalities This is my question: Is the following statement true ?
Let $H$ be a real or complex Hilbertspace and $R,S:H \to H$ compact operators.
For every $n\in\mathbb{N}$ the following inequality holds:
$$\sum_{j=1}^n s_j(RS) \leq \sum_{j=1}^n s_j(R)s_j(S)$$
Note: $s_j(R)$ denotes the j-th singular value of the opeartor $R$.
The sequence of the singular values falls monotonically to zero.
With best regards,
mat

Edit: I found out, that the statement is true for products instead of sums. By that I mean:
Let $H$ be a $\mathbb{K}$-Hilbertspace and $R,S: H \to H$ compact operators.
For every $n\in\mathbb{N}$ we have:
$$\Pi_{j=1}^n s_j(RS) \leq \Pi_{j=1}^n s_j(R)s_j(S)$$
Is it possible to derive the statement for sums from this?

The statement is true. It is a special case of a result by Horn (On the singular values of a product of completely continuous operators, Proc. Nat.Acad. Sci. USA 36 (1950) 374-375).
The result is the following. Let $f:[0,\infty)\rightarrow \mathbb{R}$ with $f(0)=0$. If $f$ becomes convex following the substitution $x=e^t$ with $-\infty\leq t<\infty$, then for any linear completely continuous operators $R$ and $S$, $$\sum_{j=1}^n f(s_j(RS))\leq \sum_{j=1}^n f(s_j(R)s_j(S)).$$ 
The function $f(x)=x$ falls in the scope of the theorem and the proof follows from the theorem you stated about products.
