If $ a\le b+c $ then $ a^q \le C(b^q + c^q) $? Let $a,b,c>0$ and $q>1$. If
$$
a\le b+c
$$
then
$$
a^q \le C(b^q + c^q)
$$
holds for some constant $C>0$?

Consider $a = 1, b = c = 1/2$. Then as $q \to \infty$, we have that your left hand side is always $1$, while the right hand side goes to $0$ (very, very, very quickly) for any $C$.
So there is no $C$ so that $a^q \leq C(b^q + c^q)$ based on the restrictions $a,b,c > 0, q > 1$. $\diamondsuit$
More broadly, for general $a$, if we let $b = c = a/2$, then we can see that no general $C$ should exist. By expanding $a$ as a binomial expansion, and noting that in this case every power $b^j c^{q-j}$ are the same, then this amounts to asking how large the binomial coefficients $q \choose k$ can get. And they can get arbitrarily large. $\diamondsuit$
