Is this Poincaré-type inequality valid? It is proved that the Poincaré inequality is still true for functions with zero mean boundary traces. Motivated by this, I have the following question:

Let $\Omega$ be an open,bounded and connected subset of $\mathbb R^3$
  with a $C^2-$boundary $\partial \Omega \equiv \Gamma$. If $f \in
 W^{1,2}(\Omega)$ then could we claim that: 
${\vert \vert f - \frac{1}{\vert \Gamma \vert } \int_{\Gamma} {\vert f \vert}^{1/2}
 \vert \vert}_{L^2(\Omega)} \leq C {\vert \vert \nabla f \vert
 \vert}_{L^2(\Omega)}$

If for any $u\in \{ u\in H^1(\Omega): \frac{1}{\vert \Gamma \vert } \int_{\Gamma} u=0 \}$ we have the estimate: ${\vert \vert u \vert \vert}_{L^2(\Omega)} \leq C {\vert \vert \nabla u \vert\vert}_{L^2(\Omega)}$ then it seems logical to me that the above claim could be true. However I wasn't able to prove it (if indeed can be proved) so any help is much appreciated.
Thanks in advance!
 A: When coming up with inequalities, it is important that the equations you get be 'dimensionally consistent'  In particular, we can often use scaling arguments to show that only certain types of inequalities are possible.  
In this particular case, let us fix a function $g$, and let $f_\lambda = \lambda g$ for all $\lambda > 0$.  Then, you inequality would would have
$$\lVert \lambda g - \sqrt{\lambda} \frac{1}{|\Gamma|} \int_{\Gamma} |g|^{1/2} \rVert_{L^2} \leq C \lambda \lVert \nabla g \rVert_{L^2}$$
If we cancel a factor of $\lambda$ from both sides, we have
$$\lVert g - \frac{1}{|\Gamma|\lambda^{1/2}} \int_\Gamma |g|^{1/2} \rVert_{L^2} \leq C \lVert \nabla g \rVert_{L^2}$$
But, as $\lambda \to 0$, $\lambda^{-1/2} \to \infty$, so the term on the left becomes larger without bound for nonzero $g$, while the term on the right is constant, which is a contradiction.

Of course, a physicist would not even need to go through this calculation: the terms $f$ and $\frac{1}{|\Gamma|} \int_{\Gamma} |f|^{1/2}$ have different units, so there's no way we could subtract them!
