Question

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Geometric-arithmetic mean 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: Using the Cauchy-Schwarz Inequality, prove that if \(a\geq 0\) and \(b\geq 0\), then \(\sqrt{ab} \leq \frac{a+b}{2}\). Answer: To prove the inequality \(\sqrt{ab} \leq \frac{a+b}{2}\), we first compute the dot product of the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) which results in \(2ab^{\frac{1}{2}}\). Next, we compute the magnitudes of both vectors and find that they are equal to \(\sqrt{a+b}\). By applying the Cauchy-Schwarz Inequality, we obtain the inequality \(|2ab^{\frac{1}{2}}| \leq a+b\). Simplifying this inequality and considering that both \(a\) and \(b\) are non-negative, we finally arrive at the inequality \(\sqrt{ab} \leq \frac{a+b}{2}\).

Step-by-step solution

Compute the dot product of the given vectors

To compute the dot product of the given vectors, we use the formula \(\mathbf{u} \cdot \mathbf{v} = u_{1}v_{1} + u_{2}v_{2}\). Therefore, \[ \mathbf{u}\cdot\mathbf{v} = \sqrt{a}\sqrt{b} + \sqrt{b}\sqrt{a} = 2ab^{\frac{1}{2}}. \]

Compute the magnitudes of the given vectors

We compute the magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\) using the formula \(|\mathbf{u}|=\sqrt{u_{1}^2+u_{2}^2}\): \[ |\mathbf{u}|=\sqrt{(\sqrt{a})^2 + (\sqrt{b})^2}=\sqrt{a+b}. \] Similarly, \[ |\mathbf{v}|=\sqrt{(\sqrt{b})^2 + (\sqrt{a})^2}=\sqrt{a+b}. \]

Apply the Cauchy-Schwarz Inequality

Now, we will use the Cauchy-Schwarz Inequality which states that \(|\mathbf{u}\cdot\mathbf{v}| \leq |\mathbf{u}||\mathbf{v}|\). Plug in the results from Steps 1 and 2: \[ |2ab^{\frac{1}{2}}| \leq |\sqrt{a+b}|\cdot|\sqrt{a+b}|. \]

Simplify the inequality

We will now simplify the inequality obtained in Step 3: \[ 2|ab^{\frac{1}{2}}| \leq a+b. \] Since both \(a\) and \(b\) are non-negative, we can divide both sides of the inequality by 2: \[ |ab^{\frac{1}{2}}| \leq \frac{a+b}{2}. \]

Conclude the proof

Since \(a\geq 0\) and \(b\geq 0\), the absolute value sign around the left side of the inequality can be removed, and the inequality becomes: \[ ab^{\frac{1}{2}}\leq\frac{a+b}{2}. \] Therefore, we have shown that \(\sqrt{ab} \leq \frac{a+b}{2}\) for \(a \geq 0\) and \(b \geq 0\).

Key concepts

Dot Product of Vectors

Understanding the dot product of vectors is essential in various fields of mathematics and physics. At its core, the dot product is a binary operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is formally defined by
\[\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + \dots + u_nv_n,\]where \(u_i\) and \(v_i\) are components of vectors \(\mathbf{u}\) and \(\mathbf{v}\), respectively. To physically interpret this, imagine the dot product as a measure of how much one vector extends in the direction of another.
The dot product also has a geometric representation:\[\mathbf{u} \cdot \mathbf{v} = |\mathbf{u}||\mathbf{v}| \cos \theta,\]where \(\theta\) is the angle between vectors \(\mathbf{u}\) and \(\mathbf{v}\), and \(|\mathbf{u}|\) and \(|\mathbf{v}|\) are the magnitudes of the vectors. The resulting dot product tells us not only about the directionality of vectors but also their relationship in terms of orthogonality when the dot product equals zero, implying that the vectors are perpendicular to each other. In our exercise, calculating the dot product of vectors \(\mathbf{u}\) and \(\mathbf{v}\) gave us a foundation to explore more complex inequalities like the Cauchy-Schwarz inequality.

Vector Magnitudes

In the realm of linear algebra, vector magnitude is a fundamental concept that plays a key role in understanding vector properties and operations. A vector's magnitude, often thought of as its 'length', is determined by the distance it spans in the n-dimensional space it occupies. Mathematically, the magnitude of a vector \(\mathbf{u}\) is calculated as:
\[|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + \dots + u_n^2},\]which stems from the Pythagorean theorem applied in n-dimensional space. It's imperative to note that the magnitude is always a non-negative value. In our exercise, you were tasked with finding the magnitudes of vectors \(\mathbf{u}\) and \(\mathbf{v}\), which both result in \(\sqrt{a+b}\), a step that later contributed to understanding the application of the Cauchy-Schwarz inequality.

Geometric-arithmetic Mean Inequality

The geometric-arithmetic mean inequality is a fundamental principle in algebra that provides a relationship between the arithmetic mean and the geometric mean of two non-negative numbers. It can be stated as follows:
\[\sqrt{ab} \leq \frac{a + b}{2},\]where \(a\) and \(b\) are non-negative real numbers. This inequality suggests that for any pair of non-negative numbers, the geometric mean (the square root of their product) will always be less than or equal to their arithmetic mean (the sum of the numbers divided by two). In the context of our exercise, we utilized this principle to demonstrate a real-world application of the Cauchy-Schwarz Inequality. Here, vectors \(\mathbf{u}\) and \(\mathbf{v}\) were carefully chosen to construct an elegant proof of the geometric-arithmetic mean inequality by relating the dot product and magnitudes of vectors within the framework provided by Cauchy-Schwarz Inequality.