# Proof involving absolute value and maximums

Given the definition for any real numbers a and b, the max function is

$$\max\{a, b\} = \begin{cases} a \text{ if } a \geq b \\ b \text{ if } a < b \\ \end{cases}$$

Lemma: for any two real numbers $$a\text{ and }b, a \leq \max\{a,b\}$$ and $$b \leq \max\{a,b\}$$.

Prove that for real numbers, $$a, b$$, and $$x$$, if a $$\leq x \leq$$ b then $$|x| \leq \max\{|a|,|b|\}$$.

So far, I have

$$x \geq 0$$ or $$x < 0$$ by the law of trichotomy

Case 1: $$x\geq 0$$

$$|x| = x$$ by the definition of absolute value

and I'm not sure where to go from there. Any hints?

## Hints Through a Conversation

I'm going to restate your lemma to make it simpler, so you can see later how this lemma is applicable to your problem.

Lemma. For all real numbers $$\alpha$$ and $$\beta$$ $$\alpha \leq \max\big(\alpha, \beta\big)$$

We don't need to state $$\beta \leq \max\big(\alpha,\beta\big)$$, because if the above lemma is true for all $$\alpha$$, then it is true for all $$\beta$$; that is $$\beta \leq \max\big(\alpha,\beta\big)$$.

## Observation

Next we need to observe the rather trivial inequality $$|\alpha| \leq |\alpha|$$. Obviously, they are equal, but it will be more useful at the moment to right it as an inequality, because, by the definition of absolute value, we obtain $$-|\alpha| \leq \alpha \leq |\alpha|$$ You can use this in your proof by saying something like "Since $$|b| \leq |b|$$ we have $$-|b| \leq b \leq |b|$$ by definition of absolute value." This is a valid move, because we are only using the definition of the absolute value and the trivial inequality $$|b|\leq|b|$$.

Now, notice that by definition of the absolute value, the inequality $$|x|\leq\max\big(|a|,|b|\big)$$ is equivalent to $$-\max\big(|a|,|b|\big)\leq x \leq \max\big(|a|,|b|\big)$$

Let's focus on the right inequality $$x \leq \max\big(|a|,|b|\big)$$. From the above lemma if we set $$\alpha = |b|$$ and $$\beta = |a|$$, then we get $$|b| \leq \max\big(|a|, |b|\big)$$ What remains to be shown is that $$x \leq |b|$$. Can you use our observation to establish the right inequality?

You now have enough information to work on a proof. I have provided my proof but concealed it. Hover over the yellowish area with your mouse to reveal my solution.

## Proof

Let $$a,b$$ and $$x$$ denote real numbers such that $$a \leq x \leq b$$. Since $$b \leq |b|$$ we have $$x \leq b \leq |b| \leq \max\big(|a|,|b|\big)$$ Further $$-\max\big(|a|, |b|\big) \leq -|a| \leq a$$, so we deduce $$-\max\big(|a|, |b|\big) \leq -|a| \leq a \leq x \leq b \leq |b| \leq \max\big(|a|, |b|\big)$$ or $$-\max\big(|a|, |b| \big) \leq x \leq \max\big(|a|, |b|\big)$$ Therefore $$|x| \leq \max\big(|a|, |b|\big)$$