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 $$\Big\lambda_1 (A+B)\lambda_1 (A)\Big^{1/3}+\Big\lambda_2 (A+B)\lambda_2 (A)\Big^{1/3}\leq\lambda_1 (B)^{1/3}+\lambda_2 (B)^{1/3}$$ for any two $2\times2$ symmetric real matrices $A$ and $B$? Thanks a lot!
closed as no longer relevant by Andrés E. Caicedo, Andy Putman, Suvrit, Will Jagy, Harald HancheOlsen Feb 21 '12 at 15:33This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question. 


The alleged inequality is false, even if you restrict $A$ and $B$ to be positive definite matrices. Consider the following, $$ A = \begin{bmatrix} 1.2281 & 0.6361\\\\ 0.6361 & 1.9690 \end{bmatrix},\quad\quad B = \begin{bmatrix} 3.7829 &0.6021\\\\ 0.6021 & 0.4002 \end{bmatrix}. $$ Then, we have the following: \begin{eqnarray*} \lambda(A+B) = (5.0114, 2.3687)\\\\ \lambda(A) = (2.3347, 0.8624)\\\\ \lambda(B) = (3.8868, 0.2962) \end{eqnarray*} From, which we see that \begin{eqnarray*}\ \ \lambda_1(A+B) \lambda_1(A)\ \ ^{1/3} + \ \ \lambda_2(A+B)\lambda_2(A)\ \ ^{1/3} & = & 2.5348\\\\ \lambda_1(B)^{1/3} + \lambda_2(B)^{1/3} &=& 2.2389 \end{eqnarray*} 

