Let $M \in \mathbb{C}^{n \times n}$ be a Hermitian matrix, then $$\sum_{i=1}^k \tilde{m}_{ii} \leq \sum_{i=1}^k \lambda_i,$$ for $k = 1,\dots,n-1$. Here $\tilde{m}_{11} \geq \tilde{m}_{22} \geq \cdots \geq \tilde{m}_{nn}$ are the diagonal elements of $M$ in nondecreasing order and $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$ are the eigenvalues of $M$ in nondecreasing order.
There are lots of posts containing some information about the Schur-Horn theorem, which I can see that my question would be a direct consequence of. However, can we prove this without appealing to this theorem? One idea I had was perhaps using min-max theorem with the standard basis vectors. For instance we know that $$\lambda_1 = \max_{x \neq 0} \frac{x^*Mx}{x^*x} \geq \frac{e_1^*Me_1}{e_1^*e_1}= m_{11},$$ (assuming that the diagonal of $M$ is already in nondecreasing order, if not we can do make $PMP^{-1}$ have the desired order for $P$ permutation). Then continue this process and sum up the terms to prove the inequality. Does this work or is something missing?
Edit: What I had in mind was the following: For $\lambda_2$, we have $$\lambda_2 = \min_{e_1,e_3,\dots,e_{n-2}} \max_{x \neq 0 \\ x \perp e_i \\ i \neq2} \frac{x^*Mx}{x^*x} \geq \min_{e_1,e_3,\dots,e_{n-2}} \frac{e_2^*Me_2}{e_2^*e_2} = m_{22}.$$ Perhaps this is not legal, but this was my interpretation of the min-max theorem.
