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 not positivedefinite matrices $A$) 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.

$\begingroup$ Where do you get these statements from? Suvrit gave a counterexample to a previous claim, so unless there is a good reason to believe this is true, it is not clear that it is worthwhile thinking about these questions. $\endgroup$– Igor RivinDec 28, 2011 at 20:52

1$\begingroup$ I think that this inequality does hold (even for $n \times n$ matrices), but I know the proof only for positive definite matrices, which implies a restricted version of the inequality that you actually seem to be after. $\endgroup$– SuvritDec 28, 2011 at 20:55

4$\begingroup$ META: tea.mathoverflow.net/discussion/1187/… $\endgroup$– Will JagyDec 29, 2011 at 6:19
1 Answer
Below I highlight that a much more general claim holds for $n\times n$ positive definite matrices, and that a slightly weaker version of your inequality holds for general symmetric matrices.
Recall a classic theorem of Ando, (T.Ando, "Comparison of norms $\f(A)f(B)\$ and $\f(AB)\$, Math. Z., 197, (1988)):
Theorem (Ando). Let $A$ and $B$ be positive semidefinite matrices, and let $\\cdot\$ be any unitarily invariant norm, and let $X = (X^TX)^{1/2}$ denote the matrix absolute value. For any nonnegative operator monotone function $f(t)$ on $[0,\infty)$,\begin{equation*} \f(A)f(B)\ \le \f(AB)\\end{equation*}
Now, in your case we can use $f(t) = t^r$ for $r \in [0,1]$, to obtain $$\A^rB^r\ \le \\ AB^r\ \,$$ which when specialized to the tracenorm (sum of singular values) yields the inequality that you desire (but for positive matrices).
This inequality immediately implies the following weaker one for general symmetric matrices $$\ f(A)  f(B) \ \le \left\f\bigl(\bigl\ AB\ \bigr\bigr)\right\,$$ which is somewhat weaker than what you desire (but may suffice for your needswhich can be elaborated upon only if you follow Igor's suggestion and tell us where you are getting these questions from, and in what context!)

$\begingroup$ In that case, I'm curious to see your proof of the positive definite case! $\endgroup$– SuvritDec 29, 2011 at 12:01

$\begingroup$ @unknown: it actually does follow because of the following theorem: $\\sigma^\arrowdown(f(A))  \sigma^\arrowdown(f(B))\ \le \f(A)f(B)\$ $\endgroup$– SuvritJan 1, 2012 at 11:08