
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
ab > ab
>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.
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.
Now for the second part of your question, we will just apply the result from the
first part. We start by letting c = ab. 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 <= ab+b
ab+b >= a
ab >= ab
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,
ad >= 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
From: Melissa Jamin 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  <= ab 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: ab <= ab AND ba <= ab
Yet another equivalent statement is obtained by replacing ab with ba, which is perfectly valid since they are equal:
Lemma 4c: ab <= ab AND ba <= ba
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.
z_{1}  z_{2} ≤ z_{1}+z_{2} ≤ z_{1} + z_{2}
Focus on the righthand side, first:
z_{1}+z_{2} ≤ z_{1} + z_{2}
Later, we will use this to prove that z_{1}  z_{2} ≤ z_{1}+z_{2} 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. z_{1} is a vector that extends from the origin to some point on the complex plane, which I will call "A". z_{2} is a vector that extends from "A" to another point I'll call "B". The vector z_{1}+z_{2} from the origin to "B" completes the triangle. If you can visualize this, then you're more than halfway 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 xaxis. Let z' be the reflection of z about the x axis. In other words, if z=x+yi, then z'=xyi.
It is useful to note that
(1) . . . z^{2} = x^{2}+y^{2} = (x+yi)(xyi) = z z'
Another useful fact is that z_{1}' z_{2}' = (z_{1} z_{2})', which can also be expressed as:
(2) . . . (z_{1}' z_{2}) = (z_{1} z_{2}')'
(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} = x^{2} + y^{2} ≥ x^{2}, and so by taking square roots of both sides,
z ≥ x(4) . . . z+z'= x+yi + xyi = 2x = 2 Real(z)
(5) . . . z_{1} z_{2} = z_{1} z_{2} = z_{1} z_{2}' = z_{1} z_{2}'
With these facts in mind, a clever proof of the triangle inequality for real numbers follows:
z_{1}+z_{2}^{2} = (z_{1}+z_{2})(z_{1}+z_{2})' . . . (by identity 1)
= (z_{1}+z_{2})(z_{1}'+z_{2}')
= z_{1} z_{1}' + z_{1} z_{2}' + z_{2} z_{1}' + z_{2} z_{2}'
= z_{1}^{2} + z_{1} z_{2}' + z_{2} z_{1}' + z_{2}^{2} . . . (by identity 1)
= z_{1}^{2} + z_{1} z_{2}' + (z_{1} z_{2}')' + z_{2}^{2} . . . (by identity 2)
= z_{1}^{2} + 2 Real( z_{1} z_{2}' ) + z_{2}^{2} . . . (by identity 4)
≤ z_{1}^{2} + 2 z_{1} z_{2}' + z_{2}^{2} . . . (by inequality 3)
= z_{1}^{2} + 2 z_{1} z_{2} + z_{2}^{2} . . . (by identity 5)
= (z_{1} + z_{2})^{2}
Take square roots of both sides to complete the proof.
We proved, above, that z_{1}+z_{2} ≤ z_{1} + z_{2}
Let z_{3}=z_{1}+z_{2}, and z_{4}=z_{2}. By the triangle inequality, proved above,
z_{3}+z_{4} ≤ z_{3} + z_{4}
z_{1}+z_{2}z_{2} ≤ z_{1}+z_{2} + z_{2}
z_{1} ≤ z_{1}+z_{2} + z_{2}
z_{1}  z_{2} ≤ z_{1}+z_{2}
Now, the proof is complete: z_{1}  z_{2} ≤ z_{1}+z_{2} ≤ z_{1} + z_{2}
Back to the Arithmetic Rules section.
Triangles gives information of a geometrical nature about triangles.
Other inequalities: CauchySchwarz, AMGM, and Chebyshev Sum Inequality
math.fullerton.edu/mathews/c2003/ComplexGeometryMod.html, by Professor John H. Matthews, California State University, Fullerton
Euclid's Elements, Book I, Proposition 20: The Triangle Inequality
The webmaster and author of this Math Help site is Graeme McRae.