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Let $\lambda_1 (\cdot)$ be the larger absolute value eigenvalue of a $2\times2$ matrix and $\lambda_2 (\cdot)$ the smaller absolute value eigenvalue of a $2\times2$ matrix, i.e. $|\lambda_1 (\cdot)| \ge |\lambda_2 (\cdot)|$. Is it true that $$\left|\left|\lambda_{1}\left(A+B\right)\right|^{1/3}-\left|\lambda_{1}\left(A\right)\right|^{1/3}\right|+\left|\left|\lambda_{2}\left(A+B\right)\right|^{1/3}-\left|\lambda_{2}\left(A\right)\right|^{1/3}\right|\leq\left|\lambda_{1}\left(B\right)\right|^{1/3}+\left|\lambda_{2}\left(B\right)\right|^{1/3}$$ for any $2\times2$ symmetric real matrix $A$ (would suffice to prove or disprove for non-positive-definite not positive-definite matrices A, by Suvrit) $A$) and $2\times2$ diagonal real matrix $B$? Thanks a lot for any helpful answers! By the way, a relevant question was answered by Suvrit here.

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Let $\lambda_1 (\cdot)$ be the larger absolute value eigenvalue of a $2\times2$ matrix and $\lambda_2 (\cdot)$ the smaller absolute value eigenvalue of a $2\times2$ matrix, i.e. $|\lambda_1 (\cdot)| \ge |\lambda_2 (\cdot)|$. Is it true that $$\left|\left|\lambda_{1}\left(A+B\right)\right|^{1/3}-\left|\lambda_{1}\left(A\right)\right|^{1/3}\right|+\left|\left|\lambda_{2}\left(A+B\right)\right|^{1/3}-\left|\lambda_{2}\left(A\right)\right|^{1/3}\right|\leq\left|\lambda_{1}\left(B\right)\right|^{1/3}+\left|\lambda_{2}\left(B\right)\right|^{1/3}$$ for any $2\times2$ symmetric real matrix $A$ (would suffice to prove or disprove for non-positive-definite matrices A, by Suvrit) and $2\times2$ diagonal real matrix $B$? Thanks a lot for any helpful answers! By the way, a relevant question was answered by Suvrit here.

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