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! 



On an eigenvalue inequality
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\*}


