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3.4 Cauchy-Schwarz Inequality and Hölder’s Inequality
We begin with the following famous theorem:
Theorem 3.4.1. (The Cauchy-Schwarz inequality) Let a 1 , · · · , a n , b 1 , · · · , b n be real numbers. Then,
(a 1 2 + · · · + a n 2 )(b 1 2 + · · · + b n 2 ) ≥ (a 1 b 1 + · · · + a n b n ) 2 .
Proof. Let A = √ a 12 + · · · + a n2 and B = √ b 1 2 + · · · + b n 2 . In the case when A = 0, we get a 1 = · · · =
a n = 0. Thus, the given inequality clearly holds. So, we may assume that A, B > 0. We may normalize to
Hence, we need to to show that
We now apply the AM-GM inequality to deduce
1 = a 1 2 + · · · + a n 2 = b 1 2 + · · · + b n 2 .
|a 1 b 1 + · · · + a n b n | ≤ 1.
|x 1 y 1 + · · · + x n y n | ≤ |x 1 y 1 | + · · · + |x n y n | ≤ x 1 2 + y 1
2
2
+ · · · + x n 2 + y n
2
2
= 1.
Exercise 13. Prove the Lagrange identity :
( n
∑
i=1
a i
2
) ( n
∑
i=1
b i
2
) ( n
) 2
∑
− a i b i =
i=1
∑
1≤i a k b k − a k b k
4
k=1 k=1
k=1
k=1
Exercise 15. ([PF], S. S. Wagner) Let a 1 , · · · , a n , b 1 , · · · , b n be real numbers. Suppose that x ∈ [0, 1].
Show that ⎛
⎞ ⎛
⎞ ⎛
⎞
n∑
⎝ a 2 i + 2x ∑ n∑
a i a j
⎠ ⎝ b 2 i + 2x ∑ n∑
b i b j
⎠ ≥ ⎝ a i b i + x ∑ 2
a i b j
⎠ .
i=1 i