Let $H$ be a Hilbert space and $K,S: H \to H$ linear self-adjoint continuous positive Operators. $K$ is additionally compact and let $\lambda_j (K), \lambda_j(KS)$ denote the Eigenvalues of the corresponding Operators and counted with their algebraic multiplicity. By $\|S\|$ we mean the Operator Norm.
Does
$$ \vert \lambda_j(KS) \vert \leq \vert \lambda_j (K) \vert \| S \|$$
hold?
My idea thus far was to try something with the min/max principle, that for positive selfadjoint compact Operator holds:
$ \lambda_j (K) = min_{V \subset H, dim(V)=j} max_{x \in V^\perp} \frac{\langle Kx,x \rangle_H}{\langle x ,x \rangle_H} = max_{x \in V_j} \frac{\langle Kx , x \rangle }{\langle x, x \rangle} $.
with $V_j$ being the $j$ dimensional Eigenspace of $K$.
The problem is that $KS$ doesn't need to be a selfadjoint therefor the min/max principle doesn't apply. The Idea now is to find a scalarproduct $\langle \cdot , \tilde{\cdot} \rangle_2$ on $H$ wich is equivalent to $\langle \cdot , \tilde{\cdot} \rangle_H$, such that $KS$ is selfadjoint. But I do not know why such a scalarproduct should exist, also I'd probably need to demand that all the Operators are positive, meaning all Eigenvalues are $\geq0 $ to use it. Wich also doesn't feel right.
With the min/max princle one ends up with (assuming KS is positive and selfadjoint and compact)
$\lambda_j(KS) = min_{V \subset H, dim(V)=j} max_{x \in V^\perp} \frac{\langle KSx,x \rangle_H}{\langle x ,x \rangle_H} \leq max_{x \in V_j^\perp} \frac{\langle KSx,x \rangle_H}{\langle x ,x \rangle_H} \leq max_{x \in V_j^\perp} \frac{\vert \vert KS x \vert \vert \vert \vert x \vert \vert}{\vert \vert x \vert \vert^2}$ $\leq max_{x \in V_j^\perp} \frac{\vert \vert Kx \vert \vert}{\vert \vert x \vert \vert} \vert \vert S \vert \vert = \lambda_j(K) \vert \vert S \vert \vert$
with $V_j$ being the $j$ dimensional Eigenspace of $K$.