Begin2.DVI

Example 6-5. (The Schwarz inequality)

Show that for any two vectors A
and B
one can write the Schwarz inequality

|A
·B
|≤| A
|| B
|the equality holding if A
and B
are colinear.

Solution If A
and B
are nonzero quantities, then |A
·B
|must be a positive quantity.

Consider a graph of the function

y=y(x) = |A
+xB
 |^2 = (A
+xB
)·(A
+xB
)

y(x) = A
·A
+xA
·B
 +xB
·A
+x^2 B
·B


y(x) = |B
|^2 x^2 + 2(A
·B
)x+|A
|^2 =ax^2 +bx +c

Note that if y(x)> 0 for all values of x, then this would imply the graph of y(x)must

not cross the x−axis. If y(x)did cross the x−axis, then the equation y(x) = 0 would

have the two roots

x=

−b±

√

b^2 − 4 ac

2 a

in which case the discriminant b^2 − 4 ac would be positive. If y(x)does not cross the

x−axis, then the discriminant would satisfy b^2 − 4 ac ≤ 0. Here b= 2(A
·B
),a=|B
|^2

and c=|A
|^2 and the condition that the discriminant be less than or equal zero can

be expressed

b^2 − 4 ac = 4(A
·B
)^2 − 4 |B
|^2 |A
|^2 ≤ 0

or |A
·B
|≤| A
|| B
|

an inequality known as the Schwarz inequality.

Example 6-6. The triangle inequality

Show that for two vectors A
and B
the inequality |A
+B
|≤|A
|+|B
|must hold.

This inequality is known as the triangle

inequality and indicates that the length of

one side of a triangle is always less than the

sum of the lengths of the other two sides.

Solution To prove the triangle inequality one can use the Schwarz inequality from

the previous example. Observe that

|A
+B
 |^2 = (A
+B
)·(A
+B
) = A
·A
+A
·B
 +B
·A
+B
·B