Cauchy-Schwarz Inequality- Statement and Proof

Proof of Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality is a cornerstone of Linear Algebra, named after mathematicians Augustin-Louis Cauchy and Hermann Schwarz. Its history dates back to the development of inequalities in Classical Algebra. The inequality is fundamental in understanding vector spaces, projections, and angles, making it invaluable in Mathematics and beyond.

Introduction

The Cauchy-Schwarz Inequality serves as a critical tool in Linear Algebra, allowing for the establishment of bounds and relationships within vector spaces. Its applications extend into Mathematics, physics, and engineering, highlighting its versatility and significance.

Statement

If \(\pmb{a_{1},a_{2},…,a_{n}}\) and \(\pmb{b_{1},b_{2},…,b_{n}}\) be all real numbers then \begin{align*} &\pmb{\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\\ &\pmb{\ge \left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}} \end{align*} the equality occurs when

\(\pmb{a_{i}=0,~i=1,2,…,n } \) or \(\pmb{b_{i}=0,~i=1,2,…,n} \) or both
\(\pmb{a_{i}=\lambda b_{i},~i=1,2,…,n} \) where \(\pmb{\lambda }\) is any non zero real number.

Proof

Let \(\pmb{a_{1},a_{2},…,a_{n}}\) and \(\pmb{b_{1},b_{2},…,b_{n}}\) be all real numbers.
To prove \begin{align} &\pmb{\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\nonumber\\ &\pmb{\ge \left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}} \end{align}

\(\fcolorbox{black}{white}{Case-1:}\)
Let \(\pmb{a_{i}=0,~i=1,2,…,n} \). Then from (1), \begin{align*} \pmb{L.H.S}=& \pmb{\left(0^{2}+0^{2}+…+0^{2}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\\ =&\pmb{0}\\ \pmb{R.H.S}=& \pmb{ \left(0.b_{1}+0.b_{2}+…+0.b_{n}\right)^{2}} \\ =&\pmb{0} \end{align*} Therefore \( \pmb{L.H.S=R.H.S}\).
\(\fcolorbox{black}{white}{Case-2:}\)
Let \(\pmb{b_{i}=0,~i=1,2,…,n} \). Then from (1), \begin{align*} \pmb{L.H.S}=& \pmb{\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(0^{2}+0^{2}+…+0^{2}\right)}\\ =&\pmb{0}\\ \pmb{R.H.S}=& \pmb{ \left(a_{1}.0+a_{2}.0+…+a_{n}.0\right)^{2}} \\ =&\pmb{0} \end{align*} Therefore \( \pmb{L.H.S=R.H.S}\).
\(\fcolorbox{black}{white}{Case-3:}\)
Let \(\pmb{a_{i}=0,~i=1,2,…,n} \) and \(\pmb{b_{i}=0,~i=1,2,…,n} \). Then from (1), \begin{align*} \pmb{L.H.S}=& \pmb{\left(0^{2}+0^{2}+…+0^{2}\right)\left(0^{2}+0^{2}+…+0^{2}\right)}\\ =&\pmb{0}\\ \pmb{R.H.S}=& \pmb{ \left(0.0+0.0+…+0.0\right)^{2}} \\ =&\pmb{0} \end{align*} Therefore \( \pmb{L.H.S=R.H.S}\).
\(\fcolorbox{black}{white}{Case-4:}\)
Let \(\pmb{a_{i}=\lambda b_{i},~i=1,2,…,n} \) where \(\pmb{\lambda }\) is any non zero real number. Then from (1), \begin{align*} \pmb{L.H.S}=& \pmb{\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\\ =&\pmb{\left(\lambda^{2}b^{2}_{1}+\lambda^{2}b^{2}_{2}+…+\lambda^{2}b^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\\ =&\pmb{\lambda^{2}\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)^{2}}\\ \pmb{R.H.S}=& \pmb{\left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}} \\ =& \pmb{\left(\lambda b^{2}_{1}+\lambda b^{2}_{2}+…+\lambda b^{2}_{n}\right)^{2}}\\ =&\pmb{\lambda^{2}\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)^{2}} \end{align*} Therefore \( \pmb{L.H.S=R.H.S}\).
\(\fcolorbox{black}{white}{Case-5:}\)
Let \(\pmb{a_{i}\ne \lambda b_{i},~i=1,2,…,n} \) where \(\pmb{\lambda }\) is any non zero real number. Then \( \pmb{a_{i}-\lambda b_{i} }\) are not all zero for any real number \(\pmb{\lambda }\). Implies \begin{align} &\pmb{\left(a_{1}-\lambda b_{1}\right)^{2}+\left(a_{2}-\lambda b_{2}\right)^{2}+…+\left(a_{n}-\lambda b_{n}\right)^{2}\gt 0}\nonumber\\ \implies & \pmb{\lambda^{2}\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)- 2\lambda\left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)}\nonumber\\ &\pmb{+\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\gt 0 } \end{align} Since every real number \(\pmb{\lambda}\) is satisfying (2), that is, there is no real number \(\pmb{\lambda}\) satisfying the equation \begin{align} &\pmb{\lambda^{2}\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)- 2\lambda\left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)}\nonumber\\ &\pmb{+\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)= 0 } \end{align} Therefore the roots of the equation (3) must be imaginary. That is, the discriminant of (3) must be negative. So, \begin{align*} &\pmb{4\left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}}\\ &\pmb{-4\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)\lt 0}\\ \implies & \pmb{\left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}}\\ &\pmb{\lt\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\\ \implies & \pmb{\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\\ &\pmb{\gt \left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}} \end{align*}

Hence, \begin{align*} &\pmb{\left(a^{2}_{1}+a^{2}_{2}+…+a^{2}_{n}\right)\left(b^{2}_{1}+b^{2}_{2}+…+b^{2}_{n}\right)}\nonumber\\ &\pmb{\ge \left(a_{1}b_{1}+a_{2}b_{2}+…+a_{n}b_{n}\right)^{2}} \end{align*}

Explanation

Example 1:
Let sequences be \(\pmb{a_{1}=1,a_{2}=2}\) and \(\pmb{b_{1}=3,b_{2}=4}\).
\( \pmb{\left(a_{1}b_{1}+a_{2}b_{2}\right)^{2} = (1 \cdot 3 + 2 \cdot 4)^2 = 121} \)
\( \pmb{\left(a_{1}^2 + a_{2}^2\right) \cdot \left(b_{1}^2 + b_{2}^2\right) = (1^2 + 2^2) \cdot (3^2 + 4^2) = 5 \cdot 25 = 125} \)
Thus, \( \pmb{121 \lt 125} \)

Example 2:
Let sequences be \(\pmb{a_{1}=-1,a_{2}=3}\) and \(\pmb{b_{1}=4,b_{2}=-2}\).
\( \pmb{\left(a_{1}b_{1} + a_{2}b_{2}\right)^{2} = (-1 \cdot 4 + 3 \cdot -2)^2 = (-10)^2 = 100} \)
\( \pmb{\left(a_{1}^2 + a_{2}^2\right) \cdot \left(b_{1}^2 + b_{2}^2\right) = (-1)^2 + 3^2) \cdot (4^2 + (-2)^2) = 10 \cdot 20 = 200} \)
Thus, \( \pmb{100 \lt 200} \)

Example 3:
Let sequences be \(\pmb{a_{1}=0,a_{2}=2}\) and \(\pmb{b_{1}=0,b_{2}=5}\).
\( \pmb{\left(a_{1}b_{1} + a_{2}b_{2}\right)^{2} = (0 \cdot 0 + 2 \cdot 5)^2 = 10^2 = 100} \)
\( \pmb{\left(a_{1}^2 + a_{2}^2\right) \cdot \left(b_{1}^2 + b_{2}^2\right) = (0^2 + 2^2) \cdot (0^2 + 5^2) = 4 \cdot 25 = 100} \)
Thus, \( \pmb{100 = 100} \)

Example 4:
Let sequences be \(\pmb{a_{1}=0,a_{2}=0}\) and \(\pmb{b_{1}=3, b_{2}=5}\).
\( \pmb{\left(a_{1}b_{1} + a_{2}b_{2}\right)^{2} = (0 \cdot 3 + 0 \cdot 5)^2 = 0^2 = 0} \)
\( \pmb{\left(a_{1}^2 + a_{2}^2\right) \cdot \left(b_{1}^2 + b_{2}^2\right) = (0^2 + 0^2) \cdot (3^2 + 5^2) = 0 \cdot 34 = 0} \)
Thus, \( \pmb{0 = 0} \)

Example 5:
Let sequences be \(\pmb{a_{1}=2,a_{2}=3}\) and \(\pmb{b_{1}=4,b_{2}=6}\).
\( \pmb{\left(a_{1}b_{1} + a_{2}b_{2}\right)^{2} = (2 \cdot 4 + 3 \cdot 6)^2 = 26^2 = 676} \)
\( \pmb{\left(a_{1}^2 + a_{2}^2\right) \cdot \left(b_{1}^2 + b_{2}^2\right) = (2^2 + 3^2) \cdot (4^2 + 6^2) = 13 \cdot 52 = 676} \)
Thus, \( \pmb{676= 676} \)

Example 6:
Let sequences be \(\pmb{a_{1}=1,a_{2}=-2,a_{3}=3}\) and \(\pmb{b_{1}=4,b_{2}=0,b_{3}=-1}\).
\( \pmb{\left(a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}\right)^{2} = (1 \cdot 4 + -2 \cdot 0 + 3 \cdot -1)^2 = 1^2 = 1} \)
\( \pmb{\left(a_{1}^2 + a_{2}^2 + a_{3}^2\right) \cdot \left(b_{1}^2 + b_{2}^2 + b_{3}^2\right) = (1^2 + (-2)^2 + 3^2) \cdot (4^2 + 0^2 + (-1)^2) = 14 \cdot 17 = 238} \)
Thus, \( \pmb{1 \lt 238} \)

Applications

Cauchy-Schwarz Inequality is extensively used in various fields of mathematics and science. It is a critical tool in proving other inequalities like the triangle inequality and Jensen’s inequality. In physics, it aids in quantum mechanics by explaining the uncertainty principle. It also has applications in statistics for analyzing covariance and correlation. Students preparing for Classical Algebra often find this inequality indispensable.

Conclusion

The Cauchy-Schwarz Inequality provides foundational insights in Linear Algebra and related fields. By exploring its proof and applications, one gains a deeper appreciation for its role in Mathematics and its practical relevance in various disciplines.

References

Weierstrass Inequality
Triangle Inequality
Jensen’s Inequality
Hölder’s Inequality
Minkowski’s Inequality

FAQs

What is the Cauchy-Schwarz Inequality?
The Cauchy-Schwarz Inequality states that the absolute value of the dot product of two vectors is less than or equal to the product of their magnitudes.
Why is the Cauchy-Schwarz Inequality important?
It is fundamental in Linear Algebra for understanding vector spaces, inner products, and projections.
How is the Cauchy-Schwarz Inequality applied in physics?
It is used to analyze relationships between physical quantities in Mathematics and physics.
Can the inequality be derived geometrically?
Yes, geometric proofs relate the inequality to angles and distances in vector spaces.
What fields utilize the Cauchy-Schwarz Inequality?
Fields such as Linear Algebra, statistics, and optimization frequently use the inequality.
What are the prerequisites to understanding the Cauchy-Schwarz Inequality?
Knowledge of vectors, dot products, and basic Classical Algebra is essential.
How does the inequality relate to equality conditions?
Equality occurs when vectors are linearly dependent.
What are examples of the Cauchy-Schwarz Inequality in statistics?
It is used to prove the Cauchy-Schwarz divergence and other statistical measures.
How does the inequality enhance computational methods?
It provides bounds that simplify calculations in Linear Algebra.
What is the connection between the Cauchy-Schwarz Inequality and Inequalities?
It is a specific type of inequality that demonstrates relationships within vector spaces.