**Note: My answer was posted before the question was edited to a different question. My counterexample still works for version 2 of the question.**

Let $\Omega=(0,2)$. Let $$\DeclareMathOperator{\dL}{d\!}u(x)=\begin{cases}0&\text{ if }x<1\\
x-1&\text{ if }x>1\end{cases}$$ 
and 
$$
v(x)=\begin{cases}x&\text{ if }x<1\\
1&\text{ if }x>1\end{cases}
$$
Then 
$$
\int_0^2 u(x)\dL x=\frac{1}{2}
$$
and
$$
\int_0^2 v(x)\dL x=\frac{3}{2}
$$
so $\bar u=\frac{1}{4}$ and $\bar v=\frac{3}{4}$. However
$$
\begin{split}
\int_0^2(u(x)-\bar u)(v(x)-\bar v)\dL x & =\int_0^1 -\frac{1}{4}\left(x-\frac{3}{4}\right)\dL x+\int_1^2 \left(x-\frac{5}{4}\right)\frac{1}{4} \dL x \\
&=\frac{1}{16}+\frac{1}{16}=\frac{1}{8}.
\end{split}
$$
But
$$
\int_0^2 |u'(x)||v'(x)|\dL x=0.
$$