An inequality about Hermitian matrices Say one knows the following statement,
That for any Hermitian matrix $H$ with eigenvalues $\lambda_1 \geq \lambda_2 ..\geq \lambda_n$ one has, that in any basis, for any positive integers $1 \leq i_1 \leq i_2 \leq...\leq i_n \leq n$ and $k$ the inequality, $\lambda_{n-(k-1)} + \lambda_{n-(k-2)} + .. + \lambda_n \leq H_{i_1 i_2} + ... + H_{i_k i_k} \leq \lambda_1 + \lambda_2 + .. + \lambda_k$


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*Then from the above does it immediately follow that in any basis the diagonal n-tuple $(H_{11},H_{22},..,H_{nn})$ for any choice of basis lies in the convex hull of the $n!$ permutations of its eigenvalue n-tuple? 

 A: I don't know if the result you want follows from the argument you have. But you may want to take a look at the so-called Schur-Horn Theorem given here. Let $H_{11},\dots,H_{nn}$ be in the decreasing order and $\lambda_i$'s in the order you have given. Define the vectors
\begin{align}
\mathbf{h}=\begin{bmatrix}H_{11} \\ \vdots \\ H_{nn}\end{bmatrix}~~,~~\mathbf{e}=\begin{bmatrix}\lambda_{1} \\ \vdots \\ \lambda_{n}\end{bmatrix}
\end{align}
Then schur-horn theorem states that (there are different variations, more famous one is related to majorization) $$\mathbf{h}=\mathbf{P}\mathbf{e}$$ where $\mathbf{P}$ is a doubly stochastic matrix (rows and columns sums to one, and its entries are non-negative)
This implies each diagonal entry of $\mathbf{H}$ is a convex combination of the eigenvalues.
It is not so hard to prove this. Say $\mathbf{H}=\mathbf{U}\Lambda\mathbf{U}^H$ be its eigen value decomposition. See if you can express $\mathbf{h}=(\mathbf{U}\circ\mathbf{U})\mathbf{e}$ where $(\circ)$ is the hadamard product. 
