Triangle Inequality

Definition 1: Absolute Value

abs(a) = a if a is nonnegative or abs(a)=-a if a is negative.

So for example:

abs(5) = 5

abs(0) = 0

abs(-1) = 1

Now, let's look at some basic properties

Lemma 1: abs(ab) = abs(a)*abs(b)

Proof:

Case I: both a,b positive

abs(ab) = ab = abs(a)*abs(b)

Case II: both a,b negative

abs(ab) = ab = (-a)*(-b) = abs(a)*abs(b)

Case III: one negative, one positive

Assume a is positive, b is negative (since a,b are symmetrical, we can switch them as necessary)

abs(ab) = -ab = a*(-b) = abs(a)*abs(b)

QED

Lemma 2: -abs(a) ≤ a ≤ abs(a)

Proof:

Case I: a is nonnegative

-a ≤ a ≤ a

so

-abs(a) ≤ a ≤ abs(a)

Case II: a is negative

a ≤ a ≤ -a

so

-abs(a) ≤ a ≤ abs(a)

QED

Lemma 3: abs(a) ≤ b if and only if -b ≤ a and a ≤ b.

Proof:

(1) Assume abs(a) ≤ b

Case I: a is nonnegative

abs(a) = a

Since abs(a) ≤ b, it follows that a ≤ b and b is nonnegative

Since b is nonnegative and a is nonnegative, then it -b ≤ a.

Case II: a is negative

Since abs(a) ≤ b, it follows that -a ≤ b which is the same as -b ≤ a and therefore b must be nonnegative.

Since b is nonnegative, it follows that a ≤ b.

(2) Assume that -b ≤ a and a ≤ b.

Case I: a is nonnegative

Since a ≤ b, it follows that b is nonnegative

So abs(a) ≤ b.

Case II: a is negative

Since -b ≤ a, it follows that b ≥ -a.

Since a is negative, -a is positive, and we have:

abs(a) ≤ b.

QED

Lemma 4: Triangle Inequality

For all real numbers a,b

abs(a + b) ≤ abs(a) + abs(b)

Proof:

(1) For all real numbers a,b (from Lemma 1 above)

-abs(a) ≤ a ≤ abs(a)

-abs(b) ≤ b ≤ abs(b)

(2) Adding these two conditions together gives us:

-[abs(a) + abs(b)] ≤ a + b ≤ abs(a) + abs(b)

(3) Let c = a+b and d =abs(a) + abs(b)

(4) Using Lemma 3, we know that:

abs(c) ≤ d if and only if -d ≤ c and c ≤ d.

(5) But using step #2, we know that:

-d = -[abs(a) + abs(b)] ≤ c = a + b

and

c = a + b ≤ d = abs(a) + abs(b)

(6) So, using step #4 we get:

abs(c) ≤ d

which is equivalent to:

abs(a + b) ≤ abs(a) + abs(b)

QED