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Let \(\mathbf{u}=\left[\begin{array}{l}{a} \\ {b}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{l}{1} \\ {1}\end{array}\right] .\) Use the Cauchy-Schwarz inequality to show that $$\left(\frac{a+b}{2}\right)^{2} \leq \frac{a^{2}+b^{2}}{2}$$

Short Answer

Expert verified
Apply Cauchy-Schwarz and rearrange: \( \left(\frac{a+b}{2}\right)^2 \leq \frac{a^2+b^2}{2} \).

Step by step solution

01

State Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality for vectors \( \mathbf{u} \) and \( \mathbf{v} \) in Euclidean space is given by: \[ (\mathbf{u} \cdot \mathbf{v})^2 \leq \|\mathbf{u}\|^2 \|\mathbf{v}\|^2 \]where \( \cdot \) denotes the dot product and \( \| \cdot \| \) denotes the norm of a vector.
02

Find Dot Product \(\mathbf{u} \cdot \mathbf{v}\)

Calculate the dot product of \( \mathbf{u} \) and \( \mathbf{v} \):\( \mathbf{u} \cdot \mathbf{v} = a \cdot 1 + b \cdot 1 = a + b \).
03

Calculate Norms

Find the norm of each vector:- The norm of \( \mathbf{u} \) is:\[ \|\mathbf{u}\| = \sqrt{a^2 + b^2} \]- The norm of \( \mathbf{v} \) is:\[ \|\mathbf{v}\| = \sqrt{1^2 + 1^2} = \sqrt{2} \]
04

Apply Cauchy-Schwarz Inequality

Substitute the values into the Cauchy-Schwarz inequality:\[ (a + b)^2 \leq (\sqrt{a^2 + b^2})^2 (\sqrt{2})^2 \]Simplify the right side:\[ (a + b)^2 \leq (a^2 + b^2) \cdot 2 \]
05

Rearrange the Inequality

Divide both sides by 4 to support the form of the desired inequality:\[ \frac{(a + b)^2}{4} \leq \frac{a^2 + b^2}{2} \]Rearranging, you get:\[ \left(\frac{a+b}{2}\right)^2 \leq \frac{a^2+b^2}{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Dot Product
The vector dot product is a fundamental operation in vector algebra. It is also known as the scalar product because it results in a single number. To compute the dot product between two vectors, say \( \mathbf{u} = [a, b] \) and \( \mathbf{v} = [1, 1] \), you multiply corresponding components of the vectors and then sum the results.

In this example, the dot product \( \mathbf{u} \cdot \mathbf{v} \) is given by:
  • Multiply the first components: \( a \cdot 1 = a \)
  • Multiply the second components: \( b \cdot 1 = b \)
  • Add the products: \( a + b \)
So, \( \mathbf{u} \cdot \mathbf{v} = a + b \). This result is pivotal in applying the Cauchy-Schwarz inequality, which is an essential concept in linear algebra.

Understanding vector dot products is key to grasping other vector operations and inequalities, making it crucial for various applications in mathematics and physics.
Euclidean Norm
The Euclidean norm of a vector, often referred to as its magnitude or length, is a measure of the "size" of the vector in Euclidean space. It is calculated using the square root of the sum of the squares of its components. For example, consider the vector \( \mathbf{u} = [a, b] \).

The Euclidean norm \( \| \mathbf{u} \| \) is computed as:
  • Square each component: \( a^2 \) and \( b^2 \)
  • Sum the squares: \( a^2 + b^2 \)
  • Take the square root: \( \sqrt{a^2 + b^2} \)
For our vectors, \( \mathbf{v} = [1, 1] \) also has a Euclidean norm:
  • Square each component: \( 1^2 = 1 \)
  • Add the squares: \( 1 + 1 = 2 \)
  • Square root: \( \sqrt{2} \)
This norm analysis is significant as it is used in the Cauchy-Schwarz inequality, which involves comparing the square of the dot product to the product of the norms. The Euclidean norm provides a way to understand vector magnitudes, which is essential in many fields of science and engineering.
Linear Inequalities
Linear inequalities are inequalities that involve a linear function. In this context, they help compare different expressions. The Cauchy-Schwarz inequality itself is a form of a linear inequality, which sets a boundary for the dot product of vectors.

For example, using vectors \( \mathbf{u} = [a, b] \) and \( \mathbf{v} = [1, 1] \), the inequality is shown by:
  • Forming the inequality: \( (a + b)^2 \leq (a^2 + b^2) \cdot 2 \)
  • Rearranging it by dividing both sides by 4: \( \frac{(a + b)^2}{4} \)
  • Showing: \( \left(\frac{a+b}{2}\right)^2 \leq \frac{a^2+b^2}{2} \)
This result indicates that the average of \( a \) and \( b \), squared, will always be less than or equal to the average of the squares of \( a \) and \( b \).

This process of rearranging and understanding inequalities is vital in mathematical proofs and problem-solving. Linear inequalities are a cornerstone of mathematics as they appear in numerous areas including optimization, economics, and statistics.

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Most popular questions from this chapter

Given \(a \geq 0\) and \(b \geq 0,\) let \(\mathbf{u}=\left[\begin{array}{c}{\sqrt{a}} \\ {\sqrt{b}}\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{c}{\sqrt{b}} \\\ {\sqrt{a}}\end{array}\right]\) Use the Cauchy-Schwarz inequality to compare the geometric mean \(\sqrt{a b}\) with the arithmetic mean \((a+b) / 2\)

In Exercises 19 and \(20,\) all vectors are in \(\mathbb{R}^{n} .\) Mark each statement True or False. Justify each answer. a. \(\mathbf{v} \cdot \mathbf{v}=\|\mathbf{v}\|^{2}\) b. For any scalar \(c, \mathbf{u} \cdot(c \mathbf{v})=c(\mathbf{u} \cdot \mathbf{v})\) c. If the distance from \(\mathbf{u}\) to \(\mathbf{v}\) equals the distance from \(\mathbf{u}\) to \(\quad-\mathbf{v},\) then \(\mathbf{u}\) and \(\mathbf{v}\) are orthogonal. d. For a square matrix \(A,\) vectors in Col \(A\) are orthogonal to vectors in Nul \(A .\) e. If vectors \(\mathbf{v}_{1}, \ldots, \mathbf{v}_{p}\) span a subspace \(W\) and if \(\mathbf{x}\) is orthogonal to each \(\mathbf{v}_{j}\) for \(j=1, \ldots, p,\) then \(\mathbf{x}\) is in \(W^{\perp}\) .

Find a formula for the least-squares solution of \(A \mathbf{x}=\mathbf{b}\) when the columns of \(A\) are orthonormal.

Involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat{\beta}\) of \(\mathbf{y}=X \beta .\) Consider the following numbers. (i) \(\|X \hat{\boldsymbol{\beta}}\|^{2}-\) the sum of the squares of the "regression term." Denote this number by \(\mathrm{SS}(\mathrm{R})\). (ii) \(\|\mathbf{y}-X \hat{\boldsymbol{\beta}}\|^{2}-\) the sum of the squares for error term. Denote this number by \(\mathrm{SS}(\mathrm{E})\). (iii) \(\|\mathbf{y}\|^{2}-\) the "total" sum of the squares of the \(y\) -values. Denote this number by \(\mathrm{SS}(\mathrm{T}) .\) Every statistics text that discusses regression and the linear model \(\mathbf{y}=X \boldsymbol{\beta}+\boldsymbol{\epsilon}\) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the \(y\) -values is zero. In this case, SS(T) is proportional to what is called the variance of the set of \(y\) -values. Justify the equation \(\mathrm{SS}(\mathrm{T})=\mathrm{SS}(\mathrm{R})+\mathrm{SS}(\mathrm{E})\). [Hint: Use a theorem, and explain why the hypotheses of the theorem are satisfied.] This equation is extremely important in statistics, both in regression theory and in the analysis of variance.

In Exercises 5 and \(6,\) describe all least-squares solutions of the equation \(A \mathbf{x}=\mathbf{b} .\) $$ A=\left[\begin{array}{lll}{1} & {1} & {0} \\ {1} & {1} & {0} \\ {1} & {0} & {1} \\ {1} & {0} & {1}\end{array}\right], \mathbf{b}=\left[\begin{array}{l}{1} \\ {3} \\ {8} \\ {2}\end{array}\right] $$

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