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 Math Help > Basic Math > Arithmetic Rules > Triangle Inequality

On 6/24/2001 10:17:16 AM, Jek Orila wrote:
>help! i'm stuck in proving the triangle inequality theorem.
>it is said that for all a and b, prove that:

|a+b| < |a|+|b|

and

|a-b| > |a|-|b|

>any help would be greatly appreciated. thanks.

I will prove these for you, assuming a and b are real numbers, starting with the first part, the sum, and then using that part to prove the second part, the difference.  At the bottom of the page, I will prove the triangle inequality for complex numbers.

Lemma 1: |a+b| <= |a| + |b|, where a and b are real numbers

First case, when a+b>=0:

a <= |a|
b <= |b|
so a+b <= |a|+|b|
If a+b >= 0 then a+b=|a+b|, and then
|a+b| <= |a|+|b|, proving this case.

Second case, when a+b < 0

If a+b < 0 then let c=-a and let d=-b.
Now we have
c <= |c|
d <= |d|
so c+d <= |c|+|d|
c+d >= 0, so c+d=|c+d|, and then
|c+d| <= |c|+|d|

|c+d| = |-(a+b)| = |a+b|
|c| = |a|
|d| = |b|

So by substituting, we get
|a+b| <= |a|+|b|, proving this case.

Lemma 2: |a-b| >= |a|-|b|, where a and b are real numbers

Now for the second part of your question, we will just apply the result from the first part. We start by letting c = a-b. We know from the first part of your question that
|c+b| <= |c|+|b|.

a=c+b, so Substituting a in place of c+b, we get
|a| <= |c|+|b|
|a| <= |a-b|+|b|
|a-b|+|b| >= |a|
|a-b| >= |a|-|b|

Lemma 3: |a+b| >= |a|-|b|, where a and b are real numbers

You didn't ask this, but another student, Peter Gibney, asked for a proof that
|a+b| >= |a| - |b|.

To prove this, let d = -b.  Starting with the result of Lemma 2,

|a-d| >= |a| - |d| , and then substitute -b in place of d, giving
|a+b| >= |a| - |d| .  Finally, substitute |b| in place of |d| (since they're equal), giving
|a+b| >= |a| - |b|

Lemma 4: | |a| - |b| | <= |a-b|, where a and b are real numbers

 From: Melissa Jamin  Sent: Wednesday, July 06, 2005 3:20 PM Subject: math help I need help proving something using the Triangle Inequality.  I can prove it using cases, but if you could help me prove it using the Triangle Inequality, I would greatly appreciate the help. The problem states: Prove | |a| - |b| | <= |a-b| using the triangle inequality. (That is, the absolute value of the difference between the absolute value of a and the absolute value of b is less than or equal to the absolute value of the difference between a and b)

An equivalent statement of Lemma 4 is this:

Lemma 4b:   |a|-|b| <= |a-b|  AND  |b|-|a| <= |a-b|

Yet another equivalent statement is obtained by replacing |a-b| with |b-a|, which is perfectly valid since they are equal:

Lemma 4c:   |a|-|b| <= |a-b|  AND  |b|-|a| <= |b-a|

Now, you will see that the two parts of Lemma 4c follow directly from Lemma 2, proving Lemma 4c as stated above.  Now, since the statement of Lemma 4c is equivalent to that of Lemma 4b, and this is equivalent to Lemma 4, it follows that Lemma 4 is also true.

Triangle Inequality for Complex Numbers

|z1| - |z2| ≤ |z1+z2| ≤ |z1| + |z2|

Focus on the right-hand side, first:

|z1+z2| ≤ |z1| + |z2|

Later, we will use this to prove that |z1| - |z2| ≤ |z1+z2| in a manner similar to Lemma 2 and Lemma 3, above.

To get yourself psyched up for this proof, think what this inequality means in geometrical terms.  z1 is a vector that extends from the origin to some point on the complex plane, which I will call "A".  z2 is a vector that extends from "A" to another point I'll call "B".  The vector z1+z2 from the origin to "B" completes the triangle.  If you can visualize this, then you're more than half-way there, because you know in your heart that the length of the third side of a triangle can't be more than the sum of the lengths of the first two sides.

There is a very clever proof of the triangle inequality that depends on some interesting properties of the reflection of a complex number about the x-axis.  Let z' be the reflection of z about the x axis.  In other words, if z=x+yi, then z'=x-yi.

It is useful to note that

(1) . . . |z|2 = x2+y2 = (x+yi)(x-yi) = z z'

Another useful fact is that z1' z2' = (z1 z2)', which can also be expressed as:

(2) . . . (z1' z2) = (z1 z2')'

(3) . . . Real(z), defined as the real component of z, is never greater than |z|, because

z = x + yi, where x and y are real numbers, and x=Real(z), so
|z|2 = x2 + y2 ≥ x2, and so by taking square roots of both sides,
|z| ≥ x

(4) . . . z+z'= x+yi + x-yi = 2x = 2 Real(z)

(5) . . . |z1 z2| = |z1| |z2| = |z1| |z2'| = |z1 z2'|

With these facts in mind, a clever proof of the triangle inequality for real numbers follows:

|z1+z2|2 = (z1+z2)(z1+z2)' . . . (by identity 1)
= (z1+z2)(z1'+z2')
= z1 z1' + z1 z2' + z2 z1' + z2 z2'
= |z1|2 + z1 z2' + z2 z1' + |z2|2 . . . (by identity 1)
= |z1|2 + z1 z2' + (z1 z2')' + |z2|2 . . . (by identity 2)
= |z1|2 + 2 Real( z1 z2' ) + |z2|2 . . . (by identity 4)
≤ |z1|2 + 2 |z1 z2'| + |z2|2 . . . (by inequality 3)
= |z1|2 + 2 |z1| |z2| + |z2|2 . . . (by identity 5)
= (|z1| + |z2|)2

Take square roots of both sides to complete the proof.

The rest of the proof, for complex numbers

We proved, above, that |z1+z2| ≤ |z1| + |z2|

Let z3=z1+z2, and z4=-z2.  By the triangle inequality, proved above,

|z3+z4| ≤ |z3| + |z4|
|z1+z2-z2| ≤ |z1+z2| + |z2|
|z1| ≤ |z1+z2| + |z2|
|z1| - |z2| ≤ |z1+z2

Now, the proof is complete: |z1| - |z2| ≤ |z1+z2| ≤ |z1| + |z2|

Related pages in this website

Back to the Arithmetic Rules section.

Triangles gives information of a geometrical nature about triangles.

Other inequalities: Cauchy-Schwarz, AM-GM, and Chebyshev Sum Inequality

Internet references

math.fullerton.edu/mathews/c2003/ComplexGeometryMod.html, by Professor John H. Matthews, California State University, Fullerton

The webmaster and author of this Math Help site is Graeme McRae.