Trace inequality for the square root of the sum of matrices Let $A$ and $B$ be two Hermitian positive semi-definite matrices.
Is it true that  $\mathrm{Tr}\sqrt{A+B}\leq \mathrm{Tr}\sqrt{A} + \mathrm{Tr}\sqrt{B}$?
The inequality is true if $A$ and $B$ commute. Does it hold in general?
 A: Yes, it's true. Assume first that $A,B$ are invertible (that is, positive definite). Then
\begin{align} 
\text {Tr}((A+B)^{1/2})
&=\text{Tr}\,((A+B)^{-1/4}(A+B)(A+B)^{-1/4})\\ \ \\
&=\text{Tr}\,((A+B)^{-1/4}A(A+B)^{-1/4})+\text{Tr}\,((A+B)^{-1/4}B(A+B)^{-1/4})\\ \ \\
&=\text{Tr}\,(A^{1/2}(A+B)^{-1/2}A^{1/2})+\text{Tr}\,(B^{1/2}(A+B)^{-1/2}B^{1/2})\\ \ \\
&\leq\text{Tr}\,(A^{1/2})+\text{Tr}\,(B^{1/2}).
\end{align}
The inequality is justified by the fact that $A\leq A+B$ and the square root is monotone, so $A^{1/2}\leq (A+B)^{1/2}$. This (nontrivially ) implies that 
$$
(A+B)^{-1/2}\leq A^{-1/2}.
$$Now conjugating with $A^{1/2}$, we get 
$$
A^{1/2}(A+B)^{-1/2}A^{1/2}\leq A^{1/2}.
$$
For the general case we get by above that, for every $\varepsilon>0$, 
$$
\text{Tr}\,((A+B+\varepsilon I)^{1/2})\leq\text{Tr}\,((A+\varepsilon I)^{1/2})+\text{Tr}\,((B+\varepsilon I)^{1/2}).
$$
The inequality then follows by making $\varepsilon\to0$. The trace is continuous, and $(A+\varepsilon I)^{1/2}-A^{1/2}=f_\varepsilon(A)$, where $f_\varepsilon(t)=(t+\varepsilon)^{1/2}-t^{1/2}$. Note that 
$$
|f_\varepsilon(t)|=\frac\varepsilon{(t+\varepsilon)^{1/2}+t^{1/2}},
$$
so $$\|(A+\varepsilon I)^{1/2}-A^{1/2}\|\leq\frac{\varepsilon}{(\lambda_m+\varepsilon)^{1/2}+\lambda_m^{1/2}}\leq\frac\varepsilon{2\lambda_m^{1/2}},
$$
where $\lambda_m$ is the smallest nonzero eigenvalue of $A$. 
