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.

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)):
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!) 

