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Let matrix $ X $ is Hermitian and denote $ \lambda_1(X) \ge \lambda_2(X) \ge \ldots \ge \lambda_n(X) $ eigenvalues of matrix $ X $.

Prove that $ \lambda_i(A + B) \le \lambda_i(A) + \lambda_1(B) $

I think that here could be applied relations that if $ A $ is Hermitian matrix and $ \lambda_1(A) \ge \lambda_2(A) \ge \ldots \ge \lambda_n(A) $ are eigenvalues of $ A $and $ B $ - principal submatrix of order $ n - 1 $ and $ \lambda_1(B) \ge \lambda_2(B) \ge \ldots \ge \lambda_{n-1}(B) $ it's eigenvalues, then $ \lambda_1(A) \ge \lambda_1(B) \ge \lambda_2(A) \ge \ldots \ge \lambda_{n-1}(B) \ge \lambda_n(A) $. Or corollaries from that theorem.

Also I tried to diagonalize matrix using Schur's theorem.

And maybe here could be useful some inequalities connected with eigenvalues, for example:

$ \sum_{i=1}^{k}\lambda_i(A+B) \le \sum_{i=1}^{k}\lambda_i(A) + \sum_{i=1}^{k}\lambda_i(B)$

Or $ \lambda_1(A+B) + \lambda_n(A+B) \le \lambda_1(A) + \lambda_n(A) + \lambda_1(B) + \lambda_2(B)$

But I do not know how to solve my problem. Thanks for the help!

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