286 CHAPTER 14. VECTOR PRODUCTS
Proof: First note that if b= 0, both sides of 14.6 equal zero and so the inequality holdsin this case. Indeed,
a ·0 = a·(0+0) = a ·0+a ·0so a ·0= 0. Therefore, it will be assumed in what follows that b ̸= 0.
Define a function of t ∈ R
f (t) = (a+ tb) · (a+ tb) .
Then by 14.2, f (t)≥ 0 for all t ∈ R. Also from 14.3,14.4,14.1, and 14.5
f (t) = a · (a+ tb)+ tb · (a+ tb) = a ·a+ t (a ·b)+ tb ·a+ t2b ·b
= |a|2 +2t (a ·b)+ |b|2 t2.
Then solve f ′ (t) = 0 for t. This gives t = −(a·b)|b|2
. Plug this value of t into the formula for
f (t). Then
0 ≤ |a|2 +2
(−(a ·b)|b|2
)(a ·b)+ |b|2
(−(a ·b)|b|2
)2
= |a|2 − 2(a ·b)2∣∣b2∣∣ +(a ·b)2
|b|2= |a|2 − (a ·b)2
|b|2= f
(−(a ·b)|b|2
)(14.7)
which shows(a ·b)2 ≤ |a|2 |b|2 , |(a ·b)| ≤ |a| |b| .
From properties of the dot product, equality holds in 14.6 whenever one of the vectorsis a scalar multiple of the other. It only remains to verify this is the only way equality canoccur. If either vector equals zero, then one is a multiple of the other. If equality holds, inthe inequality, then f (t) = 0 from 14.7. Therefore, for t the point where minimum of f isachieved, (a+ tb) · (a+ tb) = 0 and so a=−tb.
You should note that the entire argument was based only on the properties of the dotproduct listed in 14.1 - 14.5. This means that whenever something satisfies these axioms,the Cauchy Schwartz inequality holds. There are many other instances of these propertiesbesides vectors in Rp.
The Cauchy Schwartz inequality allows a proof of the triangle inequality for distancesin Rp in much the same way as the triangle inequality for the absolute value.
Theorem 14.1.6 (Triangle inequality) For a,b ∈ Rp
|a+b| ≤ |a|+ |b| (14.8)
and equality holds if and only if one of the vectors is a nonnegative scalar multiple of theother. Also
||a|− |b|| ≤ |a−b| (14.9)
Proof: By properties of the dot product and the Cauchy Schwarz inequality,
|a+b|2 = (a+b) · (a+b) = (a ·a)+(a ·b)+(b ·a)+(b ·b)
= |a|2 +2(a ·b)+ |b|2 ≤ |a|2 +2 |a ·b|+ |b|2
≤ |a|2 +2 |a| |b|+ |b|2 = (|a|+ |b|)2 .