Is that true that if A - B is positive definite, then A's eigenvalues are larger than B's

Question

Suppose that $A$ and $B$ are two Hermite matrices and that $A - B$ is positive definite. Denote the eigenvalues of these two matrices by $\lambda_1(A) \ge \lambda_2(A) \ge \dots \ge \lambda_n(A)$ and $\lambda_1(B) \ge \lambda_2(B) \ge \dots \ge \lambda_n(B)$. Then can we derive $$\lambda_i(A) > \lambda_i(B)\ \ \forall i?$$

It seems to me this is true but I don't know how to prove it. I'd appreciate any help. Thanks in advance!