Loewner order in terms of eigenvalues Suppose that $A \succeq B$, where $A$ and $B$ are real symmetric matrices, so that $A - B$ is positive semidefinite, equivalently, $A - B$ has nonnegative eigenvalues. 
Is it always true that $\lambda_i(A) \geq \lambda_i(B)$ (assuming that eigenvalues are ordered)? 

Just so that this question not remains formally unanswered: as @julian pointed out in the comments by the min-max-theorem we have that for all $k \in \{ 1, \ldots, d \}$
$$
\lambda_k(A)
= \min_{\substack{U \subset \mathbb C^d, \\ \dim(U) = k}} \max_{x \in U \setminus \{ 0 \}} \frac{x^{\mathsf{T}} A x}{x^{\mathsf{T}} x}
\ge \min_{\substack{U \subset \mathbb C^d, \\ \dim(U) = k}} \max_{x \in U \setminus \{ 0 \}} \frac{x^{\mathsf{T}} B x}{x^{\mathsf{T}} x}
= \lambda_k(B),
$$
where the inequality is due to $B \preceq A$, which is equivalent to $0 \preceq A - B$, i.e. $x^{\mathsf{T}} (A - B) x \ge 0$ for all $x \in \mathbb {C}^d$.
