<s>[INST] 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 :(
 [/INST] The difference is:
We can have $4 \ge x(x-4)$ if $x(x-4) \le 0$.
But we can not have $\frac 4{x(x-4)} \ge 1$ if $x(x-4) \le 0$.
(if $x(x-4) = 0$ then $\frac 4{x(x-4)}$ doesn't exist.  And if $x(x-4) < 0$ then $\frac 4{x(x-4)} < 0 < 1$.)
(Otherwise, when $x(x-4) > 0$ then $\frac {4}{x(x-4)} \ge 1$ and $4 \ge x(x+4)$ are the same thing.)
So $4\ge x(x-4)$ is true more often and will have more solutions than $\frac 4{x(x-4)}\ge 1$.
(Although, come to think of it,  $4 \ge x(x-4) > 0$ and $\frac 4{x(x-4)} \ge 1$ will have the exact same solutions.....)
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To solve $4 \ge x(x-4)$ we have
$x(x-4) = x^2 - 4x \le 4$ so
$x^2 - 4x + 4 \le 8$ so
$0 \le (x -2)^2 \le 8$ so
$0 \le x - 2 \le \sqrt 8$ so
$2-\sqrt 8 \le x \le \sqrt 8+2$
But notice these solutions will include cases where $x(x-4) \le 0$.
.....
TO solve $\frac 4{x(x-4)}\ge 1$ we must consider three cases:
Case 1:  $x(x-4) < 0$.
Then $\frac 4{x(x-4)} \ge 1 \implies$
$\frac 4{x(x-4)} x(x-4) \le 1\cdot x(x-4)\implies$
$4 \le x(x-4) < 0$.  But that is impossible.
Case 2: $x = 0$
Then $\frac 4{x(x-4)} = \frac 4 0$ and that is impossible.
Case 3: $x(x-4) > 0$.
Then we have either $x > 0$ and $x-4 > 0$; so $x> 4$.   Or $x < 0$ and $x-4 < 0$; so $x < 0$.  So we are restricted to requiring $x > 4$ or $x < $.  In other words $0 \le x \le 4$ is impossible.
But if we make that restriction then
$\frac 4{x(x-4)} \ge 1 \implies $
$\frac 4{x(x-4)} x(x-4) \ge 1\cdot x(x-4) \implies$
$4 \ge x(x-4)$.
But note !!!VERY IMPORTANT!!!! we can only do this because we are assuming $x(x-4) > 0$.  If we didn't know $x(x-4) > 0$ we could not say this.
Now we solved $4 \ge x(x-4)$ above and got $2-\sqrt 8 \le x \le 2 +\sqrt 8$.
But we MUST combine it with $x \not \in [0, 4]$.
So the solution here is $2-\sqrt 8 \le x < 0$ or $4 < x < 2 + \sqrt 8$.
.....
These two solutions are the same EXCEPT $4 \ge x(x-4)$ includes cases where $x(x-4) \le 0$ (which occures if and only if $0\le x \le 4$)  while  $\frac 4{x(x-4)} \ge 1$ does not include those.
[SO as the solutions to $4\ge x(x-4)$ is $[2-\sqrt 8, 2 + \sqrt 8]$ then the solutions to $\frac 4{x(x-4)} \ge 1$ is $[2-\sqrt 8, 2+\sqrt 8] \setminus [0, 4] = [2-\sqrt 8, 0) \cup (4, 2+\sqrt 8]$.
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