Question

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

Short answer

Question: Prove that for non-negative values of \(a\) and \(b\), the inequality \(\sqrt{ab} \leq (a+b)/2\) holds using the Cauchy-Schwarz Inequality and the given vectors \(\mathbf{u} = \langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v} = \langle\sqrt{b}, \sqrt{a}\rangle\). Answer: By applying the Cauchy-Schwarz Inequality to the vectors \(\mathbf{u}\) and \(\mathbf{v}\), we showed that \(2ab \leq a+b\). After dividing both sides by 2, we obtained the desired inequality: \(\sqrt{ab} \leq (a+b)/2\).

Step-by-step solution

Write down the given vectors and inequality

We know that \(\mathbf{u} = \langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v} = \langle\sqrt{b}, \sqrt{a}\rangle\). Also, we have the Cauchy-Schwarz Inequality: \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\).

Calculate the dot product of \(\mathbf{u}\) and \(\mathbf{v}\)

To find the dot product, we multiply corresponding components of the two vectors and add the results: \((\mathbf{u}\cdot\mathbf{v}) = \sqrt{a}\times\sqrt{b} + \sqrt{b}\times\sqrt{a} = ab+ab = 2ab\).

Calculate the magnitudes of \(\mathbf{u}\) and \(\mathbf{v}\)

To calculate the magnitudes, we find the square root of the sum of the squared components for each vector: \(|\mathbf{u}| = \sqrt{(\sqrt{a})^2 + (\sqrt{b})^2} = \sqrt{a+b}\) \(|\mathbf{v}| = \sqrt{(\sqrt{b})^2 + (\sqrt{a})^2} = \sqrt{b+a}\) Since \(|\mathbf{u}| = |\mathbf{v}|\), we can denote their magnitudes as \(M\), so \(M = |\mathbf{u}| = |\mathbf{v}| = \sqrt{a+b}\).

Apply the Cauchy-Schwarz Inequality to our vectors

Using the Cauchy-Schwarz Inequality with our calculated values, we have: \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\) which simplifies to: \(2ab \leq M^2\)

Substitute the expression for \(M\) and simplify

We know that \(M = \sqrt{a+b}\). Therefore, \(M^2 = a+b\). Now, substitute this expression into the inequality from Step 4: \(2ab \leq a+b\)

Divide both sides of the inequality by 2 and make the final conclusion

Dividing both sides of the inequality by 2, we get: \(\sqrt{ab} \leq\frac{a+b}{2}\) Hence, the inequality \(\sqrt{ab} \leq (a+b)/2\) is proven for non-negative values of \(a\) and \(b\).

Key concepts

Dot Product

The dot product is a fundamental operation in vector mathematics. It combines two vectors to produce a scalar value. If you have two vectors \(\mathbf{u}\) and \(\mathbf{v}\), their dot product \(\mathbf{u} \cdot \mathbf{v}\) is calculated as the sum of the products of their corresponding components. For example, if \(\mathbf{u} = \langle \sqrt{a}, \sqrt{b} \rangle\) and \(\mathbf{v} = \langle \sqrt{b}, \sqrt{a} \rangle\), the dot product is:

• \(\sqrt{a} \times \sqrt{b} + \sqrt{b} \times \sqrt{a} = ab + ab = 2ab\).

The dot product can also be expressed in terms of the angle \(\theta\) between the two vectors using the formula: \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}| |\mathbf{v}| \cos \theta\). This equation links the dot product to vector magnitudes and the cosine of the angle between them.
Understanding the dot product is crucial in proving the Cauchy-Schwarz Inequality.

Vector Magnitude

Vector magnitude, also known as the length or norm of a vector, measures how long a vector is. It is calculated as the square root of the sum of the squares of its components. For a vector \(\mathbf{u} = \langle x, y \rangle\), the magnitude \(|\mathbf{u}|\) is given by:

• \(|\mathbf{u}| = \sqrt{x^2 + y^2}\).

For our specific vectors \(\mathbf{u} = \langle \sqrt{a}, \sqrt{b} \rangle\) and \(\mathbf{v} = \langle \sqrt{b}, \sqrt{a} \rangle\), both magnitudes are equal, calculated as: \(|\mathbf{u}| = |\mathbf{v}| = \sqrt{a+b}\). This equality simplifies the application of the Cauchy-Schwarz Inequality, as shown in the solution. Vector magnitude is central to understanding many vector principles and is essential when utilizing inequalities.

Inequalities

Inequalities are mathematical statements indicating that one expression is larger or smaller than another. In this context, the Cauchy-Schwarz Inequality asserts that for any vectors \(\mathbf{u}\) and \(\mathbf{v}\):

• \(|\mathbf{u} \cdot \mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\),

which can be directly applied to prove other inequalities, such as \(\sqrt{ab} \leq \frac{a+b}{2}\).
The inequality arises from the property that the cosine of any angle is limited to the range [-1, 1]. Therefore, \(|\cos \theta| \leq 1\), and subsequently, any dot product cannot exceed the product of the magnitudes of the involved vectors. Working with inequalities in algebra often requires creative substitution and manipulation of these mathematical expressions to illustrate or solve more complex problems.

Geometric Interpretation

The geometric interpretation of vectors provides visual insight into their properties and interactions, like the Cauchy-Schwarz Inequality. On a plane, vectors have direction and magnitude. The dot product, \(\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos \theta\), helps us understand how vectors align.

• If \(\theta = 0\), the vectors point in the same direction, making \(\cos \theta = 1\).
• If \(\theta = \pi/2\), the vectors are orthogonal, leading to \(\cos \theta = 0\).

Thus, the Cauchy-Schwarz Inequality holds a significant geometric meaning: it bounds the scale of the dot product by the lengths of vectors, linked by the angle between them. Using this concept, we gain both algebraic and visual understanding of how the vectors interact in physical and theoretical spaces.