Why are the solutions for $\frac{4}{x(4-x)} \ge 1\;$ and $\;4 \ge x(4-x)\;$ different?

Question

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 :(

Answer 1

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:

$$\begin{align} {4\over x(4-x)}\ge1 &\iff{4\over x(4-x)}-1\ge0\\ &\iff{4-x(4-x)\over x(4-x)}\ge0\\ &\iff{4-4x+x^2\over x(4-x)}\ge0\\ &\iff{(2-x)^2\over x(4-x)}\ge0\\ &\iff x(4-x)\gt0\quad\text{(since }(2-x)^2\text{ is always non-negative)}\\ &\iff0\lt x\lt4 \end{align}$$

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.

Answer 2

Actually, $\frac4{x(4-x)}\geqslant1$ is equivalent to $4\geqslant x(4-x)$ if $x\in(0,4)$ since, in that case, $x(4-x)>0$.

But if $x>4$ or $x<0$, then $x(4-x)<0$, and therefore $\frac4{x(4-x)}\geqslant1\iff4\color{red}\leqslant x(4-x)$.

Answer 3

Note that when you multiply by a negative number, you need to "switch" the equal sign. A very basic example: $ -1 < 4 $ multiplied by $-1$ becomes $ 1 > -4$.

In your case you multiply both sides with $x(4-x)$ or $4x - x^2$. This is positive for $x\in(0,4)$. So this becomes:

$$ 4\leq x(4−x)\,for\,x\in[-\infty, 0] \cup [4,\infty] $$ $$ 4≥ x(4−x)\,for\,x\in[0, 4] $$

Answer 4

Suppose, for example, $x < 0$, I.e. $x$ is negative.

This would mean that $4-x>4$ and therefore $x(4-x)<0$. In other words $x(4-x)$ is negative and therefore less than or equal to $4$.

Now, you know that:

$$4 \geq x(4-x)$$

As $x(4-x)$ is negative, you also know $\frac{1}{x(4-x)}$ is negative so if you multiply the inequality above by $\frac{1}{x(4-x)}$ you have to flip the sign (as multiplying both sides of an inequality by a negative number means you have to flip the sign).

So you get:

$$\frac{4}{x(4-x)}\leq 1$$

Long story short: if you multiply an inequality by a negative number on both sides you must flip the sign around. This means when $x(4-x)<0$ the two inequalities aren't the same.

Answer 5

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 $[2-\sqrt 8, 2+\sqrt 8] \setminus [0, 4] = [2-\sqrt 8, 0) \cup (4, 2+\sqrt 8]$.

Answer 6

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 $(4-x)<0$, and in this case it is equivalent to $\dfrac4{x(4-x)}\le 1$.