Various proofs of the Cauchy-Schwarz inequality

Hui-Hua Wu and Shanhe Wu ∗

Department of Mathematics and Computer Science, Longyan University, Longyan, Fujian 364012, P. R. ChinaE-mail: wushanhe@yahoo.com.cn

∗Corresponding Author

Abstract: In this paper twelve different proofs are given for the classical Cauchy-Schwarz inequality.Keywords: Cauchy-Schwarz inequality; arithmetic-geometric means inequality; rearrangement inequality;mathematical induction; scalar product2000 Mathematics Subject Classification: 26D15

1 IntroductionThe Cauchy-Schwarz inequality is an elementary inequality and at the same time a powerful inequality,

which can be stated as follows:Theorem. Let (a1, a2, . . . , an) and (b1, b2, . . . , bn) be two sequences of real numbers, then(

n∑i=1

a2i

)(n∑

i=1

b2i

)≥

(n∑

i=1

aibi

)2

, (1)

with equality if and only if the sequences (a1, a2, . . . , an) and (b1, b2, . . . , bn) are proportional, i.e., there is aconstant λ such that ak = λbk for each k ∈ {1, 2, . . . , n}.

As is known to us, this classical inequality plays an important role in different branches of modernmathematics including Hilbert spaces theory, probability and statistics, classical real and complex analysis,numerical analysis, qualitative theory of differential equations and their applications (see [1-12]). In thispaper we show some different proofs of the Cauchy-Schwarz inequality.

2 Some different proofs of the Cauchy-Schwarz inequalityProof 1. Expanding out the brackets and collecting together identical terms we have

n∑i=1

n∑j=1

(aibj − ajbi)2 =n∑

i=1

a2i

n∑j=1

b2j +

n∑i=1

b2i

n∑j=1

a2j − 2

n∑i=1

aibi

n∑j=1

bjaj

= 2

(n∑

i=1

a2i

)(n∑

i=1

b2i

)− 2

(n∑

i=1

aibi

)2

.

Because the left-hand side of the equation is a sum of the squares of real numbers it is greater than orequal to zero, thus (

n∑i=1

a2i

)(n∑

i=1

b2i

)≥

(n∑

i=1

aibi

)2

.

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Proof 2. Consider the following quadratic polynomial

f (x) =

(n∑

i=1

a2i

)x2 − 2

(n∑

i=1

aibi

)x +

n∑i=1

b2i =

n∑i=1

(aix− bi)2.

Since f(x) ≥ 0 for any x ∈ R, it follows that the discriminant of f(x) is negative, i.e.,(n∑

i=1

aibi

)2

−

(n∑

i=1

a2i

)(n∑

i=1

b2i

)≤ 0.

The inequality (1) is proved.

Proof 3. Whenn∑

i=1

a2i = 0 or

n∑i=1

b2i =0, (1) is an identity.

We can now assume that

An =n∑

i=1

a2i 6= 0, Bn =

n∑i=1

b2i 6= 0, xi =

ai√An

, yi =bi√Bn

(i = 1, 2, . . . , n),

then

n∑i=1

x2i =

n∑i=1

y2i = 1.

The inequality (1) is equivalent to

x1y1 + x2y2 + · · ·xnyn ≤ 1,

that is

2(x1y1 + x2y2 + · · ·xnyn) ≤ x21 + x2

2 + · · ·+ x2n + y2

1 + y22 + · · ·+ y2

n,

or equivalently

(x1 − y1)2 + (x2 − y2)

2 + · · ·+ (xn − yn)2 ≥ 0,

which is evidently true. The desired conclusion follows.

Proof 4. Let A =√

a21 + a2

2 + · · ·+ a2n, B =

√b21 + b2

2 + · · ·+ b2n.

By the arithmetic-geometric means inequality, we have

n∑i=1

aibi

AB≤

n∑i=1

12

(a2

i

A2+

b2i

B2

)= 1,

so that

n∑i=1

aibi ≤ AB =√

a21 + a2

2 + · · ·+ a2n

√b21 + b2

2 + · · ·+ b2n.

Thus (n∑

i=1

aibi

)2

≤

(n∑

i=1

a2i

)(n∑

i=1

b2i

).

Proof 5. Let An = a21 + a2

2 + · · ·+ a2n, Bn = a1b1 + a2b2 + · · ·+ anbn, Cn = b2

1 + b22 + · · ·+ b2

n.It follows from the arithmetic-geometric means inequality that

AnCn

B2n

+ 1 =n∑

i=1

a2i Cn

B2n

+n∑

i=1

b2i

Cn=

n∑i=1

(a2

i Cn

B2n

+b2i

Cn

)≥ 2

n∑i=1

aibi

Bn= 2,

2

therefore

AnCn ≥ B2n,

that is

(a21 + a2

2 + · · ·+ a2n)(b2

1 + b22 + · · ·+ b2

n) ≥ (a1b1 + a2b2 + · · ·+ anbn)2.

Proof 6. Below, we prove the Cauchy-Schwarz inequality by mathematical induction.Beginning the induction at 1, the n = 1 case is trivial.Note that

(a1b1 + a2b2)2 = a2

1b21 + 2a1b1a2b2 + a2

2b22 ≤ a2

1b21 + a2

1b22 + a2

2b21 + a2

2b22 = (a2

1 + a22)(b

21 + b2

2),

which implies that the inequality (1) holds for n = 2.Assume that the inequality (1) holds for an arbitrary integer k, i.e.,(

k∑i=1

aibi

)2

≤

(k∑

i=1

a2i

)(k∑

i=1

b2i

).

Using the induction hypothesis, one has√√√√k+1∑i=1

a2i ·

√√√√k+1∑i=1

b2i =

√√√√ k∑i=1

a2i + a2

k+1 ·

√√√√ k∑i=1

b2i + b2

k+1

≥

√√√√ k∑i=1

a2i ·

√√√√ k∑i=1

b2i + |ak+1bk+1|

≥k∑

i=1

|aibi|+ |ak+1bk+1| =k+1∑i=1

|aibi|.

It means that the inequality (1) holds for n = k + 1, we thus conclude that the inequality (1) holds forall natural numbers n. This completes the proof of inequality (1).

Proof 7. Let

A = {a1b1, · · · a1bn, a2b1, · · · , a2bn, · · · , anb1, · · · anbn}B = {a1b1, · · · a1bn, a2b1, · · · , a2bn, · · · , anb1, · · · anbn}

C = {a1b1, · · · a1bn, a2b1, · · · , a2bn, · · · , anb1, · · · anbn}D = {a1b1, · · · anb1, a1b2, · · · , anb2, · · · , a1bn, · · · anbn}

It is easy to observe that the set A and B are similarly sorted, while the set C and D are mixed sorted.Applying the rearrangement inequality, we have

(a1b1)(a1b1)+ · · ·+(a1bn)(a1bn)+ (a2b1)(a2b1)+ · · ·+(a2bn)(a2bn)+ · · ·+(anb1)(anb1)+ · · ·+(anbn)(anbn)

≥ (a1b1)(a1b1)+· · ·+(a1bn)(anb1)+(a2b1)(a1b2)+· · ·+(a2bn)(anb2)+· · ·+(anb1)(a1bn)+· · ·+(anbn)(anbn),

which can be simplified to the inequality

(a21 + a2

2 + · · ·+ a2n)(b2

1 + b22 + · · ·+ b2

n) ≥ (a1b1 + a2b2 + · · ·+ anbn)2

as desired.

Proof 8. By the arithmetic-geometric means inequality, one has for λ > 0,

3

|aibi| ≤12

(λa2

i +b2i

λ

).

Choosing λ =

√n∑

i=1

b2i

/n∑

i=1

a2i in the above inequality gives

|aibi| ≤

√√√√√√√

n∑i=1

b2i

n∑i=1

a2i

a2i +

√√√√√√√n∑

i=1

a2i

n∑i=1

b2i

b2i

.

Hence

n∑i=1

|aibi| ≤12

√√√√√√√

n∑i=1

b2i

n∑i=1

a2i

n∑i=1

a2i +

√√√√√√√n∑

i=1

a2i

n∑i=1

b2i

n∑i=1

b2i

,

or equivalently

n∑i=1

|aibi| ≤12

√√√√ n∑i=1

b2i

n∑i=1

a2i +

√√√√ n∑i=1

a2i

n∑i=1

b2i

=

√√√√ n∑i=1

a2i ·

√√√√ n∑i=1

b2i .

The desired conclusion follows.

Proof 9. Construct the vectors α = (a1, a2, · · · , an) , β = (b1, b2, · · · , bn). Then for arbitrary realnumbers t, one has the following identities for scalar product:

(α + tβ) · (α + tβ) = α · α + 2 (α · β) t + (β · β)t2

⇐⇒ |α|2 + 2 (α · β) t + |β|2 t2 = |α + tβ|2 ≥ 0.

Thus

(α · β)2 − |α|2 |β|2 ≤ 0.

Using the expressions

α · β = a1b1 + a2b2 + · · ·+ anbn, |α|2 =n∑

i=1

a2i , |β|2 =

n∑i=1

b2i ,

we obtain (n∑

i=1

aibi

)2

−

(n∑

i=1

a2i

)(n∑

i=1

b2i

)≤ 0.

Proof 10. Construct the vectors α = (a1, a2, · · · , an) , β = (b1, b2, · · · , bn).From the formula for scalar product:

α · β = |α| |β| cos (α, β) ,

we deduce that

α · β ≤ |α| |β| .

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Using the expressions

α · β = a1b1 + a2b2 + · · ·+ anbn, |α|2 =n∑

i=1

a2i , |β|2 =

n∑i=1

b2i ,

we get the desired inequality (1).

Proof 11. Since the function f (x) = x2 is convex on (−∞,+∞), it follows from the Jensen’s inequalitythat

(p1x1 + p2x2 + · · ·+ pnxn)2 ≤ p1x21 + p2x

22 + · · ·+ pnx2

n, (2)where xi ∈ R, pi > 0 (i = 1, 2, . . . , n), p1 + p2 + · · ·+ pn = 1.

Case I. If bi 6= 0 for i = 1, 2, . . . , n, we apply xi = ai/bi and pi = b2i /(b2

1 + b22 + · · ·+ b2

n) to theinequality (2) to obtain that(

a1b1 + a2b2 + · · ·+ anbn

b21 + b2

2 + · · ·+ b2n

)2

≤ a21 + a2

2 + · · ·+ a2n

b21 + b2

2 + · · ·+ b2n

,

that is

(a1b1 + a2b2 + · · ·+ anbn)2 ≤ (a21 + a2

2 + · · ·+ a2n)(b2

1 + b22 + · · ·+ b2

n).

Case II. If there exists bi1 = bi2 = · · · = bik= 0, one has(

n∑i=1

aibi

)2

=

∑i 6=i1,...,ik,1≤i≤n

aibi

2

≤

∑i 6=i1,...,ik,1≤i≤n

a2i

∑i 6=i1,...,ik,1≤i≤n

b2i

≤

(n∑

i=1

a2i

)(n∑

i=1

b2i

).

This completes the proof of inequality (1).

Proof 12. Define a sequence {Sn} by

Sn = (a1b1 + a2b2 + · · ·+ anbn)2 −(a21 + a2

2 + · · ·+ a2n

) (b21 + b2

2 + · · ·+ b2n

).

Then

Sn+1 − Sn = (a1b1 + a2b2 + · · ·+ an+1bn+1)2 −

(a21 + a2

2 + · · ·+ a2n+1

) (b21 + b2

2 + · · ·+ b2n+1

)−(a1b1 + a2b2 + · · ·+ anbn)2 +

(a21 + a2

2 + · · ·+ a2n

) (b21 + b2

2 + · · ·+ b2n

),

which can be simplified to

Sn+1 − Sn = −[(a1bn+1 − b1an+1)

2 + (a2bn+1 − b2an+1)2 + · · ·+ (anbn+1 − bnan+1)

2],

soSn+1 ≤ Sn (n ∈ N).

We thus haveSn ≤ Sn−1 ≤ · · · ≤ S1 = 0,

which implies the inequality (1).

Acknowledgements. The present investigation was supported, in part, by the innovative experimentproject for university students from Fujian Province Education Department of China under Grant No.214,and, in part, by the innovative experiment project for university students from Longyan University of China.

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