# Inequality with eigenvalues

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!