As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized). Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$. Could you explain how you got the inequality? Hope we will get some clue from Seva's work.
Syang Chen
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