Vantage Math 100/V1C,V1F Triangle inequality
The triangle inequality is one of the most important identities in calculus. If you haven’t seen it in the past or don’t know the proof, then this is document is for you! I will be using it throughout the course and it comes up enough that I think you guys should see the proof.
Recall: Given a real number x we know intuitively that | x | is the size of x . More formally we can say that
| x | =
√ x 2 .
When we say that | x | < y we mean that the size of x is less than y independent of the sign of x . So | x | ≤ y is a shorter but equivalent way of saying x ≤ y and − x ≤ y.
With that bit of notation out of the way we can now state the triangle inequality.
Theorem.
(Triangle inequality) If a, b are real numbers then,
| a + b | ≤ | a | + | b | .
Proof.
Since x ≤ | x | for all x we have ab ≤ | ab | . Thus we have a
2
+ 2 ab + b
2 ≤ | a | 2
+ 2 | a || b | + | b | 2
.
Since both sides are perfect squares the above is equivalent to,
( a + b )
2 ≤ ( | a | + | b | )
2
.
By taking square roots of both sides we the triangle inequality,
| a + b | ≤ | a | + | b | .
Sometimes the following corollary is useful.
Corollary.
(Reverse triangle inequality) If a, b are real numbers then,
|| a | − | b || ≤ | a − b | .
Proof.
Note that by the triangle inequality we have,
| a | = | ( a − b ) + b | ≤ | a − b | + | b | ,
| b | = | ( b − a ) + a | ≤ | b − a | + | a | .
Or equivalently by rearranging,
| a | − | b | ≤ | a − b | and | b | − | a | ≤ | a − b | .
By the remark above this is equivalent to the reverse triangle inequality,
|| a | − | b || ≤ | a − b | .
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