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As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDIT

  Salle's answer is very instructive to me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)+Tr(A)Tr(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.

As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDIT: Salle's answer is very instructive to me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)+Tr(A)Tr(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.

EDIT II: I guess you can change the title into "A generalized Hoffman-Wielandt inequality" and add the tag "Numerical Analysis".

The Hoffman-Wielandt inequality states the following:

Let $A$ and $B$ be real symmetric matrices, $a_i$, $b_i$, $c_i$ the eigenvalues of $A$, $B$, $A+B$ respectively with $a_i\leq a_{i+1}$, etc. Then we have $(\sum_i |c_i-b_i|^2)^{1/2} \leq (\sum_i |a_i|^2)^{1/2}$.

A proof in spirit similar to Mikael's can be found in "The Algebraic Eigenvalue Problem" by Wilkinson. The $L^p$ variant can be found in a paper by Rajendra Bhatia and Ludwig. It seems here your taking the square root inside and $L^1$ norm outside somewhat make things tougher.

As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDITED ANSWER

Salle's answer is very instructive to me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ twice and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.

As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDIT

Salle's answer is very instructive to me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)+Tr(A)Tr(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.

As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDITED ANSWER

Salle's answer is very instructive for me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ twice and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.

As mentioned by Choi, the inequality is true when $A$ and $B$ commute (since $A$ and $B$ can be simultaneously diagonalized).

Using Rayleigh quotient we can see that $\left|\sqrt{\lambda_{1}\left(A+B\right)}-\sqrt{\lambda_{1}\left(B\right)}\right|\leq\sqrt{\lambda_{1}\left(A\right)}$ holds. But unfortunately the counterpart is not true for $\lambda_{2}$.

Could you explain how you got the inequality? Hope we will get some clue from Seva's work.

EDITED ANSWER

Salle's answer is very instructive to me. I would like to sketch here an elementary proof of the inequality $Tr(\sqrt{A+B})\leq Tr(\sqrt A)+ Tr(\sqrt B)$ he gave above.

Noticing that $Tr(\sqrt{A})=\sqrt{Tr(A)+2\sqrt{det(A)}}$ and $det(A+B)\leq det(A)+det(B)$, for any positive definite $A$ and $B$ in dimension $2$, one applies $\sqrt{a+b}\leq \sqrt{a}+\sqrt{b}$ twice and then get the inequality.

For your edited question, again I've only checked the case where $A$ and $B$ commute, and the answer is yes. But I failed to decipher the subtlety arises in the general case. Hope we will see a conclusive answer soon.

I guess you are considering B as a fixed vector and A a perturbation, which makes the inequality interesting.