Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different? Why is $\;\dfrac{4}{x(4-x)} \geq 1\;$ is not equal to $\;4 \geq x(4-x)\;$?

Probably a really dumb question but I just don't see it :(
Other answers (and comments) have already explained the potential pitfalls of multiplying both sides of an inequality by a quantity whose sign is sometimes negative. Here is an alternative approach that might help:
$$\begin{align}
{4\over x(4-x)}\ge1
&amp;\iff{4\over x(4-x)}-1\ge0\\
&amp;\iff{4-x(4-x)\over x(4-x)}\ge0\\
&amp;\iff{4-4x+x^2\over x(4-x)}\ge0\\
&amp;\iff{(2-x)^2\over x(4-x)}\ge0\\
&amp;\iff x(4-x)\gt0\quad\text{(since }(2-x)^2\text{ is always non-negative)}\\
&amp;\iff0\lt x\lt4
\end{align}$$
Remark: In general the quadratic in the numerator will not be a perfect square, in which case the final steps are a bit more complicated (you can wind up with more than one interval where the inequality holds). But the problem here was concocted to have a simple answer.
If $\dfrac4{x(4-x)}\ge 1$, $\:x(4-x)$ is positive, hence we can multiply both sides by $x(4-x)$ and therefore it implies $4\ge x(4-x)$.
However, the other way, $ 4\ge x(4-x)$ necessarily happens in particular if  $(4-x)&lt;0$, and in this case it is equivalent to $\dfrac4{x(4-x)}\le 1$.