6 Changed the explanation of the answer to the second question since it was not understood.

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$. 5 Corrected an error mentionned in the comments You can first (replacing$A$and$B$by$-A$or$-B$if necessary) assume that$Tr(A),Tr(B)\geq 0$It involves some computation. Denote by$\alpha_1\geq \alpha_2$the eigenvalues of$A$, and the same with$\beta$for$B$, and$\gamma$for$A+B$. By the positive trace assumptions,$\alpha_i=\lambda_i(A)$and so on. The But$\lambda_2(A+B)$is$\gamma_2$or$\gamma_1$, depending on the sign of$Tr(A+B)$.The possible values of for$\gamma_2$\lambda_2(A+B)$thus form an interval $I$, and on the endpoints of this interval correspond to the case when $A$ and $B$ commute.

Now, you can study how $\sum_i \left|\sqrt{|\gamma_i|} - \sqrt{|\beta_i|}\right|$ the LHS of your inequality varies as $\gamma_2$ \lambda_2(A+B)$varies, and (unless I did a mistake), . Since$|\lambda_2(A+B)| \leq |\lambda_1(A+B)|$, you see get that its the derivative of the LHS has the same sign as the derivative of$|\sqrt{|\lambda_2(A+B)|} - \sqrt{|\lambda_2(B)|}|$. Its only local maxima are maximum on the interior of$I$therefore corresponds to$\lambda_2(A+B)=0$if$0\in I$. The global maximum of the LHS on$I$is therefore reached at the endpoints of$I$, and or at$\gamma_2=0$\lambda_2(A+B)=0$ if $0\in I$.

$\left|\sqrt{Tr(A+B)} \left|\sqrt{|Tr(A+B)|} - \sqrt{|\beta_1|}\right|+\sqrt{|\beta_2|} sqrt{|\lambda_1(B)|}\right|+\sqrt{|\lambda_2(B)|} \leq \sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ provided that $0$ belongs to $I$. This Calculus should be decidable then allow you to find the maximum of the LHS on the domain defined by the inequalities $0 \in finite timeI$, $A$ being fixed. Then I expect that checking whether this maximum is less then $\sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ should be again elementary calculus.

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EDIT: I do not have the time to check the details, but the following should lead to an answer to your second question.

You can first (replacing $A$ and $B$ by $-A$ or $-B$ if necessary) assume that $Tr(A),Tr(B)\geq 0$.

Denote by $\alpha_1\geq \alpha_2$ the eigenvalues of $A$, and the same with $\beta$ for $B$, and $\gamma$ for $A+B$. By the positive trace assumptions, $\alpha_i=\lambda_i(A)$ and so on.

Given $\alpha$ and $\beta$, the possible values for $\gamma$ are described by Horn's inequalities. For $2 \times 2$ matrices, they are described by$$\gamma_1 = Tr A + Tr B - \gamma_2,$$$$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$The possible values of $\gamma_2$ form an interval $I$, and on the endpoints of this interval correspond to the case when $A$ and $B$ commute.

Now, you can study how $\sum_i \left|\sqrt{|\gamma_i|} - \sqrt{|\beta_i|}\right|$ varies as $\gamma_2$ varies, and (unless I did a mistake), you see that its local maxima are the endpoints of $I$, and $\gamma_2=0$ if $0\in I$.

But you know that on the endpoints, the inequality is verified since the matrices then commute. You are thus left to see whether$\left|\sqrt{Tr(A+B)} - \sqrt{|\beta_1|}\right|+\sqrt{|\beta_2|} \leq \sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ provided that $0$ belongs to $I$. This should be decidable in finite time.

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