72
1 Preliminaries
then there is
n
k=1
a
2
k = 1,
n
k=1
b
2
k = 1, according the Young inequality or inequality
(a k − b k )
2
≥ 0, there is
a k b k ≤
a
2
k + b
2
k
2
Sum the above inequality, we obtain
n
k=1
a k b k ≤
n
k=1
a
2
k
2
+
n
k=1
b
2
k
2
=
1
2
+
1
2
= 1
a k and b k are expressed as x k and y k respectively, there is
n
k=1
a k b k =
n
k=1
x k y k
n
i=1
x
2
i
1
2
n
i=1
y
2
i
1
2
≤ 1
or
n
k=1
x k y k ≤
n
k=1
x
2
k
1
2
n
k=1
y
2
k
1
2
When the two series on the right side of the above expression converge, let n → ∞,
the inequality (1.8.6) can be obtained. Quod erat demonstrandum.
Square the two ends of Eq. (1.8.6), the Cauchy inequality can also be written in
the following form
∞
k=1
x k y k
2
≤
∞
k=1
x
2
k
∞
k=1
y
2
k
(1.8.7)
Let R be a real number field, arbitrary real numbers x k , y k , z k ∈ R, k = 1, 2, · · · ,
make use of the Cauchy inequality, then there is
n
k=1
(x k − y k ) 2 =
n
k=1
[(x k − z k ) + (z k − y k )] 2
=
n
k=1
(x k − z k )[(x k − z k ) + (z k − y k )] +
n
k=1
(z k − y k )[(x k − z k ) + (z k − y k )]
=
n
k=1
(x k − z k )(x k − y k ) +
n
k=1
(z k − y k )(x k − y k )