Triangle inequality

The triangle inequality takes a few common forms, for example: [math]d(x,z)\le d(x,y)+d(y,z)[/math] (see metric space) of which [math]|x-z|\le|x-y|+|y-z|[/math] is a special case.

Another common way of writing it is [math]|a+b|\le |a|+|b|[/math], notice if we set [ilmath]a=x-y[/ilmath] and [ilmath]b=y-z[/ilmath] then we get [math]|x-y+y-z|\le|x-y|+|y-z|[/math] which is just [math]|x-z|\le|x-y|+|y-z|[/math]

Definition

The triangle inequality is as follows:

• [math]|a+b|\le |a|+|b|[/math]

Proof

We have 4 cases:

1. Suppose that [ilmath]a>0[/ilmath] and [ilmath]b>0[/ilmath]

   We see immediately that [ilmath]a>0\implies a+b>0+b=b>0[/ilmath] so [ilmath]a+b>0[/ilmath]

   • thus [ilmath]\vert a+b\vert=a+b[/ilmath]

   We also see that [ilmath]\vert a\vert=a[/ilmath] as [ilmath]a>0[/ilmath]

   and that [ilmath]\vert b\vert=b[/ilmath] for the same reason.

   • thus [ilmath]\vert a\vert+\vert b\vert=a+b[/ilmath]

   • We see that [ilmath]\vert a+b\vert=\vert a\vert+\vert b\vert[/ilmath]
     • Notice [ilmath]\vert a+b\vert=\vert a\vert+\vert b\vert\implies\vert a+b\vert\le\vert a\vert+\vert b\vert[/ilmath] in the literal sense of "if left then right"

       ([ilmath]\implies[/ilmath] denotes logical implication)

2. Suppose that [ilmath]a>0[/ilmath] and [ilmath]b\le 0[/ilmath]
3. Suppose that [ilmath]a\le 0[/ilmath] and [ilmath]b> 0[/ilmath]
   • Mathematicians are lazy, as [ilmath]\mathbb{R} [/ilmath] is a field (an instance of a ring) we know that [ilmath]a+b=b+a[/ilmath] (addition is commutative)
   • As [ilmath]\vert\cdot\vert:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] is a function we know that if [ilmath]x=y[/ilmath] then [ilmath]\vert x\vert=\vert y\vert[/ilmath]
   • As, again, [ilmath]\mathbb{R} [/ilmath] is a ring, we know that [ilmath]\vert a\vert+\vert b\vert=\vert b\vert+\vert a\vert[/ilmath]
   • So:
     • [ilmath]\vert a+b\vert[/ilmath]

       [ilmath]=\vert b+a\vert[/ilmath] by the commutativity of addition on [ilmath]\mathbb{R} [/ilmath]

       [ilmath]\le \vert b\vert+\vert a\vert[/ilmath] by the [ilmath]2^\text{nd} [/ilmath] case (above)

       [ilmath]=\vert a\vert+\vert b\vert[/ilmath] again by commutativity of the real numbers

   • Thus [ilmath]\vert a+b\vert\le \vert a\vert+\vert b\vert[/ilmath]
4. Both [ilmath]a\le 0[/ilmath] and [ilmath]b\le 0[/ilmath]

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TODO: Finish proof

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Reverse Triangle Inequality

This is [math]|a|-|b|\le|a-b|[/math]

Proof

Take [math]|a|=|(a-b)+b|[/math] then by the triangle inequality above:
[math]|(a-b)+b|\le|a-b|+|b|[/math] then [math]|a|\le|a-b|+|b|[/math] clearly [math]|a|-|b|\le|a-b|[/math] as promised

Note

However we see [math]|b|-|a|\le|b-a|[/math] but [math]|b-a|=|(-1)(a-b)|=|-1||a-b|=|a-b|[/math] thus [math]|b|-|a|\le|a-b|[/math] also.

That is both:

• [math]|a|-|b|\le|a-b|[/math]
• [math]|b|-|a|\le|a-b|[/math]

Full form

There is a "full form" of the reverse triangle inequality, it combines the above and looks like: [math]|a-b|\ge|\ |a|-|b|\ |[/math]

It follows from the properties of absolute value, I don't like this form, I prefer just "swapping" the order of things in the abs value and applying the same result