When trying to prove $\forall u\in W^{1,p}(U)$, $1\leq p<n$: $$\|u-u_U\|_{L^q(U)}\leq C\|Du\|_{L^p(U)}$$ where $q=\frac{np}{n-p}$, and $U$ is bounded, open with $C^1$ boundary.
I came across this answer:
(They have $U=B_r$ the $r$-ball)
Hint: First prove that $$\|u-\overline{u}_{B_r}\|_p\le c\|\nabla u\|_p,\ \forall u\in W^{1,p}(B_r)\tag{1}.$$
You can prove $(1)$ by contradiction and you will need Rellich-Kondrachov theorem. To conclude, you can use Sobolev inequality: $$\|u\|_{p^\star}\le c\|u\|_{1,p},\ \forall\ u \in W^{1,p}(B_r),$$
which will imply that $$\|u-\overline{u}_{B_r}\|_{p^\star}\le c(\|u-\overline{u}_{B_r}\|_p+\|\nabla u\|_p)\le c\|\nabla u\|_p,\ \forall\ u\in W^{1,p}(B_r).$$
Now try to match the constants and you will reach the desired inequality.
Should the last line have, one of these two corrections?
$$\|u-\overline{u}_{B_r}\|_{p^\star}\le c(\|u-\overline{u}_{B_r}\|_p+\|\nabla u\|_p)\le \color{red}{(c^2+c)}\|\nabla u\|_p,\ \forall\ u\in W^{1,p}(B_r).$$
Or
$$\|u-\bar{u}_{B_r}\|_{p^\star}\leq c\|u-\bar{u}_{B_r}\|_p + \|\nabla(u\color{red}{-\bar{u}_{B_r}})\|_p$$
The gradient of a constant is zero, no need to insert $\nabla \bar{u}_{B_r}$. In PDE theory it is customary to denote by $c$ or $C$ a constant that may assume different values in the same chain of inequalities. It is a slight abuse of notation, but it would be cumbersome to label each constant like $c_1$, $c_2$,..., $c_{104}$, and so on.