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 $

  • 2
    $\begingroup$ The reason you need $0 < x < 4$ is so that $x(4 - x)$ is positive. Between lines 5 and 6 of your solution, you divide by $x(4 - x)$, which would swap the direction of the inequality if $x(4 - x)$ was negative. $\endgroup$ Jun 18, 2022 at 23:26
  • $\begingroup$ The issue you need to worry about is not that $(x-2)^2 \ge 0$. You do know that absolutely is always true because all squares are non-negative. And your proof does follow fine up to $4\ge x(4-x)$. That is always true for any possible $x$. It's true for huge positive $x$. ($4 \ge 1568(4-1568)=1568\cdot(-1564)=-1568\cdot1564$). It is true for small positives ($4\ge 3(4-3)=3$) and it's true for all negatives ($4\ge -2(4-(-2))=-2\cdot 6=-12$). The issue is the next step when you divide by $x(4-x)$ and not flip the $\ge$ sign. You can only do that if $x(4-x)>0$. So... to be cont. $\endgroup$
    – fleablood
    Jun 19, 2022 at 1:21
  • $\begingroup$ Essentially the core of you proof is to prove that $0< x < 4\implies x(4-x) > 0$. That's the key thing, and it is the one thing you have left out of your proof. $\endgroup$
    – fleablood
    Jun 19, 2022 at 1:23
  • $\begingroup$ Hi @fleablood, I was taking that into account when dividing by $ x(4-x) $ and not changing the equality sign. Doing that I'm creating a constraint there of x not 0, x not 4 and x(4-x) > 0. Wouldn't that be correct? $\endgroup$
    – Jon
    Jun 19, 2022 at 8:48
  • 1
    $\begingroup$ I'm not attempting to prove $x-2 > 0$. That's not true. I'm just trying to say: If for some reason you think you are not allowed to assume $M^2 \ge 0$ for all $M$ then.... just prove $(x-2)^2 \ge 0$. Just reinvent the wheel. There are three cases: Either 1) $x-2 > 0$, if so $(x-2)^2 > 0$ and we are done. or 2) $x-2 = 0$, if so then $(x-2)^2 = 0$ and we are done. or 3) $x-2 < 0$. If so then $2-x > 0$ and $(x-2)^2 = (2-x)^2 > 0$ and we are done. We just re-proved the $(x-2)^2 \ge 0$ for all $x$. ... Of course we didn't NEED to do that. But if we did, we could. $\endgroup$
    – fleablood
    Jun 19, 2022 at 17:38

3 Answers 3


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 $0 < x < 4$.


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 $x<4$.

$\frac 4{x(4-x)} \ge 1$.


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 $0 < x < 4.$

$$\text{HM} \leq \text{AM} \Rightarrow \frac2{\frac1x + \frac1{4-x}} \leq \frac{x + (4-x)}2 = 2$$


$$\frac1x + \frac1{4-x} = \frac{4}{x(4-x)} \geq 1$$


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