Let $A = A^T \in \mathbb{R}^{n \times n}$ and $B = B^T \in \mathbb{R}^{n \times n}$ be any symmetric matrices. Then, I know that if $A$ and $B$ are positive definite, and $A-B$ is positive semi-definite, that is, $A, B>0$ and $A \geq B$, then \begin{equation} \lambda_j(A) \geq \lambda_j(B) > 0 \;\; \forall j =1,2,\cdots,n, \end{equation} where $\lambda_j(X)$ for a symmetric matrix $X = X^T \in \mathbb{R}^{n \times n}$ denotes the $j$-th eigenvalue of $X$ in a decreasing order: $\lambda_n(X) \leq \cdots \leq \lambda_1(X)$.
Regarding this, I considered the following converse statement.
$\qquad \qquad \lambda_j(A) \geq \lambda_j(B) > 0$ $\forall j =1,2,\cdots,n$ $\quad\Longrightarrow \quad $ $A \geq B$.
Now, I have found the simple counter example of this converse statement:
$\qquad\qquad A = \begin{bmatrix} 2 & 0 \\ 0 & 1\end{bmatrix}$, $\qquad B = \begin{bmatrix} 0.5 & 0 \\ 0 & 1.5 \end{bmatrix}$.
In this case, $\lambda_1(A) = 2 \geq 1.5 = \lambda_1(B) > 0$ and $\lambda_2(A) = 1 \geq 0.5 = \lambda_2(B) > 0$ hold, but \begin{equation} A - B = \begin{bmatrix} 1.5 & 0 \\ 0 & -0.5 \end{bmatrix}, \end{equation} so $A-B$ is indefinite. The question is: "what are the other conditions required for $A \geq B$ in addition to the eigenvalue inequalities $\lambda_j(A) \geq \lambda_j(B) > 0$?"
Unless the smallest eigenvalue of $A$ is $\ge $ largest eigenvalue of $B$ you can do the same trick with diagonal matrices that you did above. Hence the condition is : $\lambda_n(A) \ge \lambda_1(B)$.