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The answer is yes.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

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

Given $\alpha$ and $\beta$, the possible values for $\gamma$ are described by Horn's inequalities. For $2 \times 2$ matrices, they are described by $$\gamma_1 = Tr A + Tr B - \gamma_2,$$ $$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$ But $\lambda_2(A+B)$ is $\gamma_2$ or $\gamma_1$, depending on the sign of $Tr(A+B)$.The possible values for $\lambda_2(A+B)$ thus form an interval $I$, and the endpoints of this interval correspond to the case when $A$ and $B$ commute.

Now, you can study how the LHS of your inequality varies as $\lambda_2(A+B)$ varies. Since $|\lambda_2(A+B)| \leq |\lambda_1(A+B)|$, you get that 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 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 $I$, or at $\lambda_2(A+B)=0$ 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 $\left|\sqrt{|Tr(A+B)|} - \sqrt{|\lambda_1(B)|}\right|+\sqrt{|\lambda_2(B)|} \leq \sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ provided that $0$ belongs to $I$. Calculus should then allow you to find the maximum of the LHS on the domain defined by the inequalities $0 \in I$, $A$ being fixed. Then I expect that checking whether this maximum is less then $\sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ should be again elementary calculus.

The answer to both your questions are yes. Let me start with the first question, which more straightforward.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

EDIT: It seems from the comments that my answer to your second question was far from clear. Let me try to explain differently the proof I had in mind.

For 6 real numbers $\alpha_1 \geq \alpha_2$, $\beta_1\geq \beta_2$ and $\gamma_1 \geq \gamma_2$, denote by $f(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$ the quantity $\sqrt{|\alpha_1|} + \sqrt{|\alpha_2|} - |\sqrt{\max(|\gamma_1|,|\gamma_2|)} - \sqrt{\max(|\beta_1|,|\beta_2|)}| - |\sqrt{\min(|\gamma_1|,|\gamma_2|)} - \sqrt{\min(|\beta_1|,|\beta_2|)}|$.

You are asking whether $f \geq 0$ provided that $\alpha,\beta,\gamma$ are the ordered eigenvalues of respectively $A,B,A+B$ for symmetric $2 \times 2$ matrices $A$ and $B$. The answer is yes, and I am sketching a proof. Denote by $D$ the possible values for $(\alpha_1,\alpha_2,\beta_1 , \beta_2,\gamma_1,\gamma_2)$.

$D$ is exactly described by Horn's inequalities. These inequalities are $$\alpha_1 \geq \alpha_2 \ \ , \ \ \beta_1\geq \beta_2,$$ $$\gamma_1 + \gamma_2= \alpha_1 + \alpha_2+\beta_1+\beta_2,$$ $$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$

In particular, $D$ is a convex subset of dimension $5$ of $\mathbb R^6$, and one easily checks that its boundary corresponds to the case when $A$ and $B$ commute. Since the inequality is true when $A$ and $B$ commute (this is eay to check, see the other answer), your question reduces to whether $\inf_D f = \inf_{\partial D} f$. This transforms your eigenvalue question to a purely calculus question.

Notice now that $\beta,\gamma$ and $\alpha_1+\alpha_2$ being fixed, $f(\alpha,\beta,\gamma)$ decreases as $\min(|\alpha_1|,|\alpha_2|)$ decreases. Moreover, if you started with $\alpha,\beta,\gamma$ in the interior of $D$, you stay in $D$ if you make $\min(|\alpha_1|,|\alpha_2|)$ decrease, until you reach the boundary of $D$, or $\min(|\alpha_1|,|\alpha_2|)=0$. You are therefore left to prove that $f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in D$ with $\min(|\alpha_1|,|\alpha_2|)=0$.

In the same way, fixing $\alpha,\beta$ and $\gamma_1+\gamma_2$, you reduce the question to proving that $f(\alpha,\beta,\gamma) \geq \inf_{\partial D} f$ if $(\alpha,\beta,\gamma) \in D$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$.

Last, fixing $\alpha, \gamma$ and $\beta_1+\beta_2$ with $\min(|\alpha_1|,|\alpha_2|)=0$ and $\min(|\gamma_1|,|\gamma_2|)=0$, you see that $f(\alpha,\beta,\gamma)$ decreases as $\min(|\beta_1|,|\beta_2|)$ increases, until you reach the boundary of $D$. This proves that $\inf_D f = \inf_{\partial D} f$.

The answer is yes.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

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.

The answer is yes.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

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

Given $\alpha$ and $\beta$, the possible values for $\gamma$ are described by Horn's inequalities. For $2 \times 2$ matrices, they are described by $$\gamma_1 = Tr A + Tr B - \gamma_2,$$ $$\alpha_2+\beta_2 \leq\gamma_2 \leq \min(\alpha_1+\beta_2,\alpha_2+\beta_1).$$ But $\lambda_2(A+B)$ is $\gamma_2$ or $\gamma_1$, depending on the sign of $Tr(A+B)$.The possible values for $\lambda_2(A+B)$ thus form an interval $I$, and the endpoints of this interval correspond to the case when $A$ and $B$ commute.

Now, you can study how the LHS of your inequality varies as $\lambda_2(A+B)$ varies. Since $|\lambda_2(A+B)| \leq |\lambda_1(A+B)|$, you get that 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 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 $I$, or at $\lambda_2(A+B)=0$ 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 $\left|\sqrt{|Tr(A+B)|} - \sqrt{|\lambda_1(B)|}\right|+\sqrt{|\lambda_2(B)|} \leq \sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ provided that $0$ belongs to $I$. Calculus should then allow you to find the maximum of the LHS on the domain defined by the inequalities $0 \in I$, $A$ being fixed. Then I expect that checking whether this maximum is less then $\sqrt{|\alpha_1|}+\sqrt{|\alpha_2|}$ should be again elementary calculus.

The answer is yes.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

The answer is yes.

A first remark: since $A$ is positive-definite, $\lambda_i(A+B) \geq \lambda_i(B)$ for $i=1,2$. (to check this, use the formulas $\lambda_1(X) = \max_{\xi} \langle X\xi,\xi\rangle$ and $\lambda_2(X) = \min_{\xi} \langle X\xi,\xi\rangle$ where the min and max run over all unit vectors $\xi$. This formulas hold whenever $X$ is a symmetric $2 \times 2$ matrix.)

Your question is therefore whether $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ for any symmetric positive definite matrices $A$ and $B$, or equivalently $Tr( \sqrt{X X^{T} + Y Y^{T} } ) \leq Tr(\sqrt{X X^{T} })+Tr(\sqrt{Y Y^{T} })$ for any matrices $X$ and $Y$, where $X^{T}$ denotes the transpose of $X$, or hermitian tranpose if you work with complex matrices. This inequality is true in any dimension (not just 2), and it is just the triangle inequality for the Schatten 1-norm given by $\|X\|_1 = Tr(\sqrt{X X^{T} })$. The expression $Tr(\sqrt{X X^{T} + Y Y^{T} })$ is indeed the 1-norm of the matrix $\begin{pmatrix}X&Y \\\\ 0&0\end{pmatrix}$.

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.