Prove that $ \frac{4}{x(4-x)} \geq 1 $ if $ 0 < x < 4 $ From the Book of Proof, Chapter 4, exercise 12, I need to prove that $ \frac{4}{x(4-x)} \geq 1 $ if $ 0 < x < 4 $ where x is a real number. Unfortunately, even-numbered exercises don't offer a solution so I don't know if my approach is correct.
The proof I wrote is: We know that any real number squared is greater or equal to zero. We choose a real number (x-2) within the interval $ 0 < x < 4 $.
$$
(x-2)^2 \geq 0 \\
x^2-4x+4 \geq 0 \\
4 \geq 4x-x^2 \\
4 \geq x(4-x) \\
\frac{4}{x(4-x)} \geq 1
$$
My main doubt with the proof is the initial assumption that:

We know that any real number squared is greater or equal to zero. We choose a real number (x-2) within the interval $ 0 < x < 4 $

As I don't know if I can use this knowledge for the proof. Otherwise I stuck on how to start from the given information that $ 0 < x < 4 $ and reach $ (x-2)^2 \geq 0 $
 A: What you have is correct, and you did use the information in the last step.  When you divided by $x(4-x)$, to preserve the sense of the inequality, you must have $x(4-x)>0$, which is equivalent to $0 < x < 4$.
A: You can absolutely assume $(x-2)^2 \ge 0$ because $M^2 \ge 0$ always.
And the first four lines of your proof are always true.
$(x-2)^2 \ge 0$.  That is always true.
$x^2 -4x + 4 \ge 0$.  That is always true.
$4 \ge 4x -x^2= x(4-x)$.  That is always true.
No admittedly that does look like it ought to always be true but it is!  And you just proved it!!!  (Notice that $y=x(4-x)$ is a parabola that opens downward.  For $x < 0$ and $x > 4$ the values are negative and the hit maximum value when $x=2$ of $y = 4$)
So $4 \ge x(4-x)$ is ALWAYS true.
Our concern about what range of $x$ can be in only becomes an issue on the next step when you divide by $x(4-x)$ and you don't "flip the inequality".  This is only possible if $x(4-x)> 0$.  If $x(4-x) = 0$ you can't divide at all.  And if $x(4-x) < 0$ then you'd have to flip the inequality when you divide.
So that is the issue of why we have to take $0< x < 4$ into consideration.  We have to also prove that if $0 < x< 4 \implies x(x-4) > 0$.  That is the only part of your proof you didn't address.
But that part is easy.  If $0 < x < 4$ then $x > 0$ and $4-x > 0$ so $x(x-4) > 0$.
Now your proof can go like this.
$(x-2)^2 \ge 0$.  This is always true.
$x^2 -4x + 4 \ge 0$.  This is always true.
$4 \ge 4x - x^2 = x(4-x)$. This is always true.
$4 \ge x(4-x) > 0$.  This is NOT always true but it is true if $x >0$ and $x<4$.
$\frac 4{x(4-x)} \ge 1$.
A: What you have is right, but I thought I would contribute the obligatory AM-GM-HM inequality solution: consider the means of $x$ and $4-x$, which are both positive for $0 < x < 4.$
$$\text{HM} \leq \text{AM} \Rightarrow \frac2{\frac1x + \frac1{4-x}} \leq \frac{x + (4-x)}2 = 2$$
yielding
$$\frac1x + \frac1{4-x} = \frac{4}{x(4-x)} \geq 1$$
