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 $C>0$?

Let $g \colon (0,\infty) \to \mathbb{R}$ be defined by $g(x) = x^q$. Notice that $g$ is a convex function for $q > 1$, indeed $g''(x) = q(q - 1)x^{q - 2}$ which is positive.
This gives $$\Big(\frac{b + c}{2}\Big)^q = g\Big(\frac{b + c}{2}\Big) \le \frac{g(b) + g(c)}{2} = \frac{b^q + c^q}{2}.$$
Then $$a^q \le (b + c)^q \le 2^{q - 1}(b^q + c^q).$$
