Poincaré constant of a cover If $U\subset \mathbb{R}^n$ is a bounded, open, connected set and $U \subseteq \bigcup_{i=1}^N{U_i}$ (with, say, $U_i$ open bounded and connected), and $C_V$ denotes the $L^2(V)$-Poincarè constant of $U$ (that is, the best constant $C$ realizing the inequality $||u - \frac{1}{|V|}\int_V u||_{L^2(V)} \le C||\nabla u||_{L^2(V)}$ for every $u \in W^{1, 2}(V)$), is it true an inequality like
$$
 C_U \le \sum_{i=1}^N{C_{U_i}}
$$
? I suspect this is not true, but I was not able to find any counterexample. Any help would be really appreciated. 
 A: The idea of counterexample is to take two domains such that their union has a narrow bottleneck. For example (in two dimensions):  
$$V_1=\{(x,y): 0<x<1, \ 0<y<x^2\}$$
$$V_2=\{(x,y): -1<x<0, \ 0<y<x^2\}$$
Well, these are actually disjoint, so let's push them closer together: 
$$U_1 = V_1-\delta  e_x,\quad U_2 = V_2+\delta  e_x$$
where $e_x$ is the vector $(1,0)$. 
Note that translation does not change the Poincaré  constant. But the Poincaré  constant of the union $U=U_1\cup U_2$  blows up as $\delta\to 0$. Indeed, let 
$$u(x,y) = 
\begin{cases} -1,\quad &x\le -\delta \\ 
1,\quad &x\ge \delta \\ 
\delta^{-1}x,\quad & |x|\le \delta
\end{cases}
$$ 
Then the $L^2$ norm of $u$ is about the same for all $\delta$, but the $L^2$ norm of $\nabla u$ tends to zero: $\int |\nabla u|^2$  involves integrating $\delta^{-2}$ over a region of area $O(\delta^3)$.

After posting the above, I noticed you did not require $U$ to be equal to the union, only to be contained in it. Then a counterexample can be much simpler: let $U_1$ be the unit ball, and $U$ some domain contained in it, with a narrow  bottleneck of some sort. 
