The answer is to both your questions are yes. Let me start with the first question, which more straightforward.
EDIT: I do not have the time to check the details, but It seems from the following should lead to an comments that my answer to your second question was far from clear. It involves some computationLet me try to explain differently the proof I had in mind.
Denote by
For 6 real numbers $\alpha_1\geq \alpha_1 \alpha_2$ the eigenvalues of geq \alpha_2$, $A$, \beta_1\geq \beta_2$ and $\gamma_1 \geq \gamma_2$, denote by $f(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$ the same with quantity $\beta$ for \sqrt{|\alpha_1|} + \sqrt{|\alpha_2|} - |\sqrt{\max(|\gamma_1|,|\gamma_2|)} - \sqrt{\max(|\beta_1|,|\beta_2|)}| - |\sqrt{\min(|\gamma_1|,|\gamma_2|)} - \sqrt{\min(|\beta_1|,|\beta_2|)}|$.
You are asking whether $B$, and f \geq 0$ provided that $\gamma$ \alpha,\beta,\gamma$ are the ordered eigenvalues of respectively $A,B,A+B$ for symmetric $A+B$.
Given 2 \times 2$ matrices $\alpha$ A$ and $\beta$, B$. The answer is yes, and I am sketching a proof. Denote by $D$ the possible values for $\gamma$ are (\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$.
$D$ is exactly described by Horn's inequalities. For These inequalities are$2 $\alpha_1 \times 2$ matricesgeq \alpha_2 \ \ , they are described by\ \ \beta_1\geq \beta_2,$$$$\gamma_1 = Tr A + Tr B - \gamma_2,$$gamma_2= \alpha_1 + \alpha_2+\beta_1+\beta_2,$$$$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$But
In particular, $\lambda_2(A+B)$ D$ is $\gamma_2$ or $\gamma_1$, depending on the sign a convex subset of $Tr(A+B)$.The possible values for dimension $\lambda_2(A+B)$thus form an interval 5$ of $I$, \mathbb R^6$, and the endpoints of this interval correspond one easily checks that its boundary corresponds to the case when $A$ and $B$ commute.
Now, you can study how Since the LHS of your inequality varies as is true when $\lambda_2(A+B)$ varies. Since A$ and $|\lambda_2(A+B)| \leq |\lambda_1(A+B)|$, you get that the derivative of the LHS has the same sign as B$ commute (this is eay to check, see the derivative of other answer), your question reduces to whether $|\sqrt{|\lambda_2(A+B)|} - \inf_D f = \sqrt{|\lambda_2(B)|}|$. Its only local maximum on the interior of $I$ therefore corresponds inf_{\partial D} f$. This transforms your eigenvalue question to a purely calculus question.
Notice now that $\lambda_2(A+B)=0$ if \beta,\gamma$ and $0\in I$\alpha_1+\alpha_2$ being fixed, $f(\alpha,\beta,\gamma)$ decreases as $\min(|\alpha_1|,|\alpha_2|)$ decreases. The global maximum of the LHS on Moreover, if you started with $I$ is therefore reached at \alpha,\beta,\gamma$ in the endpoints interior of $I$, or at D$, you stay in $\lambda_2(A+B)=0$ D$ if $0\in I$.
But you know that on the endpointsmake $\min(|\alpha_1|,|\alpha_2|)$ decrease, the inequality is verified since until you reach the matrices then commute. boundary of $D$, or $\min(|\alpha_1|,|\alpha_2|)=0$. You are thus therefore left to see whetherprove that $\left|\sqrt{|Tr(A+B)|} - f(\alpha,\beta,\gamma) \sqrt{|\lambda_1(B)|}\right|+\sqrt{|\lambda_2(B)|} geq \leq inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ provided that in D$ with $0$ belongs to \min(|\alpha_1|,|\alpha_2|)=0$.
In the same way, fixing $I$. Calculus should then allow \alpha,\beta$ and $\gamma_1+\gamma_2$, you to find the maximum of the LHS on the domain defined by reduce the inequalities question to proving that $0 f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in I$D$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$.
Last, fixing $A$ being fixed. Then I expect \alpha, \gamma$ and $\beta_1+\beta_2$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$, you see that checking whether this maximum is less then $\sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ should be again elementary calculusf(\alpha,\beta,\gamma)$ decreases as $\min(|\beta_1|,|\beta_2|)$ increases, until you reach the boundary of $D$. This proves that $\inf_D f = \inf_{\partial D} f$.
