Are the eigenvalues of the sum of two positive definite matrices increased?

Let $A$ and $B$ be two $n \times n$ (symmetric) positive definite matrices, and denote the $k$th smallest eigenvalue of a general $n \times n$ matrix by $\lambda_k(X)$, $k = 1, 2, \ldots, n$ so that $$\lambda_1(X) \leq \lambda_2(X) \leq \cdots \leq \lambda_n(X).$$ I guess the following relation holds: $$\lambda_k(A + B) > \max\{\lambda_k(A), \lambda_k(B)\}, \; k = 1, 2, \ldots, n.$$

This looks intuitive but I have difficulty to prove it, any hints?

For symmetric matrices you have the Courant-Fischer min-max Theorem: $$\lambda_k(A) = \min \{ \max \{ R_A(x) \mid x \in U \text{ and } x \neq 0 \} \mid \dim(U)=k \}$$ with $$R_A(x) = \frac{(Ax, x)}{(x,x)}.$$ Now, your assertion follows easily, since $R_{A+B}(x) > \max\{R_A(x), R_B(x)\}$.

This theorem is also helpful to prove other nice properties of the eigenvalues of symmetric matrices. For example: \begin{equation*} \lambda_k(A) + \lambda_1(B) \le \lambda_k(A+B) \le \lambda_k(A) + \lambda_n(B) \end{equation*} This shows the continuous dependence of the Eigenvalues on the entries of the matrix, and also your assertion.