Breaking Down Complex Inequalities | Algebra

Sometimes, our given inequalities involve multiple signs such as $a < b < c$ . How do we handle these complex inequalities? To put it simply – one at a time. This is nothing but two inequalities put together so we can pull them apart for our ease and convert them to $a < b$ and $b < c$ . That said, it is not necessary to break them up. The principles of a regular inequality apply to this too.

Given $a < b < c$ ,

we can say that $a + 1 < b + 1 < c + 1$

we can also say that $a - 4 < b - 4 < c - 4$

also $2a < 2b < 2c$

and also $-a > -b > -c$

You get the point …

Let’s look at a question to see how we can split it into two and work on it.

Recall that, given $a < b$ , $(x - a)(x - b) < 0$ gives us the range $a < x < b$ and $(x - a)(x - b) > 0$ gives us the range $x < a$ or $x > b$ .

Question: If ${x}$ is positive, which of the following could be the correct ordering of ${\frac{1}{x}}$ , ${2x}$ and ${x^2}$ ?

(I) ${x^2 < 2x < \frac{1}{x}}$

(II) ${x^2 < \frac{1}{x} < 2x}$

(III) ${2x < x^2 < \frac{1}{x}}$

(A) none

(B) I only

(D) I and II

(E) I, II and III

Solution: The question has three complex inequalities. We will take each in turn. Note that each inequality consists of two more inequalities. We will split the complex inequality into two simpler inequalities e.g. $x^2 < 2x < \frac{1}{x}$ gives us $x^2 < 2x$ and $2x < \frac{1}{x}$ . Next we will find the range of values of $x$ which satisfy each of these two inequalities and we will see if the two ranges have an overlap i.e. whether there are any values of $x$ which satisfy both these simpler inequalities. If there are, it means there are values of $x$ which satisfy the entire complex inequality too. Things will become clearer once we start working on it so hold on.

Let’s look at each inequality in turn. We start with the first one:

(I) $x^2 < 2x < \frac{1}{x}$

We split it into two inequalities:

(i) $x^2 < 2x$

We can rewrite $x^2 < 2x$ as $x^2 - 2x < 0$ or $x(x - 2) < 0$ .

We know the range of $x$ for such inequalities can be easily found using the curve on the number line. This will give us $0 < x < 2$ .

(ii) $2x < \frac{1}{x}$

It can be rewritten as $x^2 - \frac{1}{2} < 0$ (Note that since $x$ must be positive, we can easily multiply both sides of the inequality with $x$ )

This gives us the range $\frac{-1}{\sqrt2} < x < \frac{1}{\sqrt2}$ (which is $0 < x < \frac{1}{\sqrt2}$ since $x$ must be positive).
Is there a region of overlap in these two ranges i.e. can both inequalities hold simultaneously for some values of $x$ ? Yes, they can hold for $0 < x < \frac{1}{\sqrt2}$ . Hence, $x^2 < 2x < \frac{1}{x}$ will be true for the range $0 < x < \frac{1}{\sqrt2}$ . So this could be the correct ordering. Let’s go on to the next complex inequality.

(II) $x^2 < \frac{1}{x} < 2x$

Again, let’s break up the inequality into two parts:

(i) $x^2 < \frac{1}{x}$

$x^1 < \frac{1}{x}$ is rewritten as $x^3 - 1 < 0$ which gives us $x < 1$ .

(ii) $\frac{1}{x} < 2x$

$\frac{1}{x} < 2x$ is rewritten as $x^2 - \frac{1}{2} > 0$ which gives us $x < \frac{-1}{\sqrt2}$ (not possible since $x$ must be positive) or $x > \frac{1}{\sqrt2}$

Can both $x < 1$ and $x > \frac{1}{\sqrt2}$ hold simultaneously? Sure! For $\frac{1}{\sqrt2} < x < 1$ , both inequalities will hold and hence $x^2 < \frac{1}{x} < 2x$ will be true. So this could be the correct ordering too.