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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 $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$

Note: My answer was posted before the question was edited to a different question. My counterexample still works for edit 6 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 $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$

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 $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$

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 $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$

Let $\Omega=(0,2)$. Let $$u(x)=\begin{cases}0&\text{ if }x<1\\ x-1&\text{ if }x>1\end{cases}$$ 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 $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$

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 $$ \int_0^2 |u'(x)||v'(x)|\dL x=0. $$