For positive numbers $a$ and $b$ we have the inequality $\sqrt{a+b} \leqslant \sqrt{a} + \sqrt{b}$. Is it true that the same holds if we take $a$ and $b$ to be positive semidefinite matrices?


If not, there is a weaker statement that I am interested in:
Is it true that the inequality $\sigma\_n( \sqrt{A} - \sqrt{B}) \leqslant \sigma\_n(|A-B|^{\frac{1}{2}})$, where by $\sigma\_n$ I denote the $n$-th singular value (they are listed in decreasing order), holds?


If even this fails, maybe the following is true: $\sigma\_{2n}(\sqrt{A} - \sqrt{B}) \leqslant \sigma\_{n} (|A-B|^{\frac{1}{2}})$?


I suspect that analogous inequalities should hold for any $0<r<1$.



Subadditivity of the square root for matrices
The claim is false. Just try some random psd matrices $A$ and $B$. You can get $\sigma\_n( |A-B|^{1/2}) = \sigma\_n^{1/2}(A-B) = 0$, whereas $\sigma\_n(A^{1/2}-B^{1/2}) > 0$.


Here is an explicit example:


\begin{equation\*}
 A = \begin{pmatrix} 19 & 17 & 9\\
 17 & 17 & 11\\
 9 & 11 & 11\end{pmatrix},\quad
 B = \begin{pmatrix}19 & 11 & 21\\
 11 & 9 & 15\\
 21 & 15 & 27\end{pmatrix},\quad A-B = 
\begin{pmatrix}0 & 6 & -12\\
 6 & 8 & -4\\
 -12 & -4 & -16\end{pmatrix}.
\end{equation\*}
Now, $\sigma\_n(\sqrt{A}-\sqrt{B}) = 0.1853...$, while $\sigma\_n(|A-B|^{1/2})=0$.


However, a weaker claim that holds is described [in this MO post](https://mathoverflow.net/questions/87872/ratio-sum-comparison-on-operators/87961#87961), namely a *weak majorization* relation:


\begin{equation\*}
 \|f(A) - f(B)\| \le \|f(|A-B|)\|,
\end{equation\*}
for any symmetric (i.e., unitarily invariant) norm $\|\cdot\|$ and
where $f(t) = t^r$, for $0< r < 1$, and more generally, $f$ is a nonnegative concave function.


