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 15:33
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2$\begingroup$ You might give a link to the m.se question so we could easily check what's there and not duplicate work. $\endgroup$ – Gerry Myerson Dec 28 '11 at 2:10

2$\begingroup$ Where does this question come from? $\endgroup$ – Igor Rivin Dec 28 '11 at 11:59

$\begingroup$ @unknown (yahoo): my sincere apologies for misreading your new question in haste. [earlier comment now deleted] (But the restriction that $B$ be diagonal real, rather than merely realsymmetric, still seems superfluous.) $\endgroup$ – Yemon Choi Dec 28 '11 at 20:30

$\begingroup$ In fact, I suggest you post the new question as a separate question, but include a link back to this one in order to provide background context $\endgroup$ – Yemon Choi Dec 28 '11 at 20:32
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*}

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