An elementary inequality of operators Suppose $a,b$ are two positive-definite linear operators on (say) $\mathbb R^n$. For $p\in(0,1)$, do we then have $(a+b)^p\leq a^p+b^p$ (with respect to the Loewner order)?
 A: Iosif's example can be given a more conceptual description. Take $a=P$, $b=Q$ as projections. Then $P^p=P$, $Q^p=Q$, so the desired inequality becomes
$$
(P+Q)^p \le P+Q .
$$
Now $T^p\le T$, for $0<p<1$, is equivalent to $T$ having no eigenvalues in $(0,1)$.
However, it's easy to give $P+Q$ an eigenvalue in this range.
For example, if $R(P),R(Q)$ span the whole space and there is a $v$ with $Pv=0$, $Qv\not= v$, then the smallest eigenvalue lies in $(0,1)$, by min-max.
A: No. E.g., (identifying linear operators with matrices in a standard manner) let
$$a=\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 0 \\
\end{array}
\right),\quad
b=\frac12 
\left(
\begin{array}{cc}
 1 & 1 \\
 1 & 1 \\
\end{array}
\right).
$$
Then (by straightforward but somewhat tedious calculations) for all $p\in(0,1)$
$$d(p):=\det(a^p+b^p-(a+b)^p) \\ 
=2^{-p-3/2} \\
\times\left(\sqrt{2} \left(2^p+2\right)-2 \left(\sqrt{2}-1\right)
   \left(2+\sqrt{2}\right)^p-2 \left(1+\sqrt{2}\right) \left(2-\sqrt{2}\right)^p\right)<0 \tag{1}\label{1},$$
and hence $(a+b)^p\not\le a^p+b^p$.

To prove the inequality in \eqref{1}, let
$$d_0(p):=d(p)2^{p+3/2},\quad 
d_1(p):=\frac{d_0'(p)}{(2+\sqrt2)^p},\\ 
d_2(p):=d_1'(p)(1+1/\sqrt2)^p. 
$$
Then
$$d_2'(p)=\left(1+\sqrt{2}\right) 2^{1-p} \left(2-\sqrt{2}\right)^p \ln\left(3-2
   \sqrt{2}\right) \ln\left(2-\sqrt{2}\right) \ln\left(2+\sqrt{2}\right)>0,$$
$d_2$ is increasing (on $(0,1)$), $d_2(1)=\sqrt{2} \ln\left(2-\sqrt{2}\right) \ln\left(6+4 \sqrt{2}\right)<0$, $d_2<0$, $d_1$ is decreasing, $d_1(1)=0$, $d_1>0$, $d_0$ is increasing, $d_0(1)=0$, $d_0<0$, $d<0$. $\quad\Box$

Here is the graph $\{(p,d(p)(1-p)^{-2})\colon0<p<1\}$:

