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# Triangle inequality

schedule Aug 12, 2023
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Theorem.

# Triangle inequality

For all $a\in\mathbb{R}$ and $b\in\mathbb{R}$, the following is true:

$$\vert{a+b}\vert\le\vert{a}\vert +\vert{b}\vert$$

This means that the absolute value of the sum of two real numbers is always less than or equal to the sum of their absolute values.

Proof. Let $a\in\mathbb{R}$ and $b\in\mathbb{R}$. By definition of absolute values, the following is true:

$$$$\label{eq:PCwFwysSWTt37PSocwN} a\le\vert{a}\vert,\;\;\;\;\; b\le\vert{b}\vert$$$$

For example:

• if $a$ is a positive number like $5$, then $5\le5$ is true.

• if $a$ is a negative number like $-6$, then $-6\le\vert{-6}\vert$ is true.

Adding the two inequalities in \eqref{eq:PCwFwysSWTt37PSocwN} gives:

$$$$\label{eq:JVs1Ho9a1hqDZPIbwNR} a+b\le\vert{a}\vert+\vert{b}\vert$$$$

The right-hand side of \eqref{eq:JVs1Ho9a1hqDZPIbwNR} matches the right-hand side of the triangle inequality. We will be using this inequality shortly.

Let's now consider the left-hand side of the triangle inequality. $\vert{a+b}\vert$ can be expressed as:

$$\vert{a+b}\vert =\begin{cases} a+b,&\text{if }a+b\ge0\\ -(a+b),&\text{if }a+b\lt0\\ \end{cases}$$

In the 1st case, if $a+b\ge0$, then $a+b=\vert{a+b}\vert$. Substituting this into \eqref{eq:JVs1Ho9a1hqDZPIbwNR} gives us the desired result:

$$$$\label{eq:euHyRrhU6V9zurpJsKG} \vert{a+b}\vert \le\vert{a}\vert+\vert{b}\vert$$$$

Next, we consider the 2nd case when $a+b\lt0$, which implies $-(a+b)=\vert{a+b}\vert$. Let's simplify this:

\label{eq:xVaeOu1ZFGfxeZu8M8N} \begin{aligned}[b] -(a+b)&=\vert{a+b}\vert\\ (-a)+(-b)&=\vert{a+b}\vert\\ \end{aligned}

We now use the property of absolute values \eqref{eq:PCwFwysSWTt37PSocwN} to obtain the following inequalities:

$$$$\label{eq:f3TgfKyiiGEy9i4oub6} -a\le\vert{-a}\vert=\vert{a}\vert,\;\;\;\;\; -b\le\vert{-b}\vert=\vert{b}\vert$$$$

Adding the two inequalities in \eqref{eq:f3TgfKyiiGEy9i4oub6} gives:

$$$$\label{eq:d1SA5UD9w1ytPHI7icz} (-a)+(-b)\le \vert{a}\vert+ \vert{b}\vert$$$$

Applying inequality \eqref{eq:d1SA5UD9w1ytPHI7icz} to \eqref{eq:xVaeOu1ZFGfxeZu8M8N} gives us the desired result:

$$\vert{a+b}\vert \le\vert{a}\vert+ \vert{b}\vert$$

This completes the proof.

Example.

## Intuition behind triangle inequality

We've proven the triangle inequality but let's now plug in some numbers to see that it indeed holds. The triangle inequality states that:

$$\vert{a+b}\vert\le\vert{a}\vert +\vert{b}\vert$$

If $a=5$ and $b=-8$, then:

$$\vert{5-8}\vert\le \vert{5}\vert +\vert{-8}\vert \;\;\;\;\; \Longleftrightarrow \;\;\;\;\; 3\le 13$$

Great, we see that the triangle inequality holds!

Theorem.

# Variant of the triangle inequality

If $a\in\mathbb{R}$ and $b\in\mathbb{R}$, then:

$$\vert{a-b}\vert \ge \vert{a}\vert- \vert{b}\vert$$

This is also equivalent to the following:

$$\vert{b-a}\vert \ge \vert{b}\vert- \vert{a}\vert$$

Proof. Let $a\in\mathbb{R}$ and $b\in\mathbb{R}$. The triangle inequality states that:

$$$$\label{eq:VYcedHjjGcYKY93viHt} \vert{a+b}\vert\le\vert{a}\vert +\vert{b}\vert$$$$

Let $c=a+b$, which implies $a=c-b$. Substituting these values into \eqref{eq:VYcedHjjGcYKY93viHt} gives:

$$\vert{c}\vert \le\vert{c-b}\vert +\vert{b}\vert$$

Rearranging this gives:

$$\vert{c-b}\vert\ge \vert{c}\vert- \vert{b}\vert$$

Since $b\in\mathbb{R}$ and $c\in\mathbb{R}$, this completes the proof of the first part.

We now assume that the following holds:

$$\vert{a-b}\vert \ge \vert{a}\vert- \vert{b}\vert$$

This must hold for any values $a$ and $b$ in $\mathbb{R}$. This means that we can simply swap the positions of $a$ and $b$, that is:

$$\vert{b-a}\vert \ge \vert{b}\vert- \vert{a}\vert$$

This completes the proof but let's take a moment to demonstrate why this is true. For example, let $a=5$ and $b=3$. The following two inequalities must hold:

\begin{align*} \vert{3-5}\vert\ge\vert{5}\vert-\vert3\vert\\ \vert{3-5}\vert\ge\vert{3}\vert-\vert5\vert\\ \end{align*}

Similarly, consider the second case when $a=3$ and $b=5$. The following must hold:

\begin{align*} \vert{5-3}\vert\ge\vert{3}\vert-\vert5\vert\\ \vert{5-3}\vert\ge\vert{5}\vert-\vert3\vert\\ \end{align*}

We see that in both cases, the theorem holds true.

Theorem.

# Reverse triangle inequality

For all $a,b\in\mathbb{R}$, we have that:

$$\Big\vert \vert{a}\vert- \vert{b}\vert \Big\vert\le \vert{a-b}\vert$$

Proof. We know from the variant triangle inequality that the following inequalities hold:

\label{eq:GFhehdE3ZtK40DVOqWx} \begin{aligned} \vert{a-b}\vert \ge\vert{a}\vert- \vert{b}\vert\\ \vert{b-a}\vert \ge\vert{b}\vert- \vert{a}\vert \end{aligned}

Since $\vert{a-b}\vert=\vert{b-a}\vert$, the second inequality can be written as:

$$\vert{a-b}\vert \ge\vert{b}\vert- \vert{a}\vert$$

Multiplying by negative one to both sides gives:

$$$$\label{eq:kCW8dpPaYpdFGzAH6Pf} -\vert{a-b}\vert \le\vert{a}\vert- \vert{b}\vert$$$$

Combining the top inequality of \eqref{eq:GFhehdE3ZtK40DVOqWx} and \eqref{eq:kCW8dpPaYpdFGzAH6Pf} gives:

$$$$\label{eq:L66ZIfBJviauRNngafv} -\vert{a-b}\vert \le\vert{a}\vert- \vert{b}\vert\le \vert{a-b}\vert$$$$

Now, recall the following basic property of absolute values:

$$-x \le{y}\le x \;\;\;\;\Longleftrightarrow\;\;\;\; \vert{y}\vert \le{x}$$

Where $x$ and $y$ are any value in $\mathbb{R}$. Applying this to \eqref{eq:L66ZIfBJviauRNngafv} gives us the desired result:

$$\Big\vert\vert{a}\vert- \vert{b}\vert\Big\vert\le \vert{a-b}\vert$$

This completes the proof.

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