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 $$\int_0^2 u(x)dx=\frac{1}{2}$$
and
$$\int_0^2 v(x)dx=\frac{3}{2}$$
so $\bar u=\frac{1}{4}$ and $\bar v=\frac{3}{4}$. However

$$\int_0^2(u(x)-\bar u)(v(x)-\bar v)dx=\int_0^1 -\frac{1}{4}(x-\frac{3}4)dx+\int_1^2 (x-\frac{5}{4})\frac{1}4 dx=\frac{1}{16}+\frac{1}{16}=\frac{1}{8}.$$

But

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