Isoif'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.

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.