The Cauchy-Schwarz Inequality Examples 1

# The Cauchy-Schwarz Inequality Examples 1

Recall from The Cauchy-Schwarz Inequality that if $V$ is an inner product space then for all vectors $u, v \in V$ we have that:

(1)
\begin{align} \quad \mid <u, v> \mid ≤ \| u \| \| v \| \end{align}

Equality in the inequality above holds if and only if one of $u$ or $v$ is a scalar multiple of one another.

We will now look at some examples regarding the Cauchy-Schwarz inequality.

## Example 1

Using the Cauchy-Schwarz inequality, prove that $\biggr | \int_{-1}^{1} p(x) q(x) \: dx \biggr |^2 ≤ \left ( \int_{-1}^{1} (p(x))^2 \: dx \right ) \left ( \int_{-1}^{1} (q(x))^2 \: dx \right )$.

Let $V$ be the set of continuous and real-valued functions defined on the interval $[-1, 1]$. Define the inner product on $V$ to be:

(2)
\begin{align} \quad <p(x), q(x)> = \int_{-1}^{1} p(x) q(x) \: dx \end{align}

We note that:

(3)
\begin{align} \quad \| p(x) \| = \sqrt{ <p(x), p(x)>} = \sqrt{\int_{-1}^{1} (p(x))^2 \: dx} \end{align}
(4)
\begin{align} \quad \| q(x) \| = \sqrt{ <q(x), q(x)>} = \sqrt{\int_{-1}^{1} (q(x))^2 \: dx} \end{align}

Therefore by the Cauchy-Schwarz inequality we have that:

(5)
\begin{align} \quad \biggr | \int_{-1}^1 p(x) q(x) \: dx \biggr | ≤ \left ( \sqrt{\int_{-1}^{1} (p(x))^2 \: dx} \right ) \left ( \sqrt{\int_{-1}^{1} (q(x))^2 \: dx} \right ) \end{align}

Both sides of the inequality above are positive ,and when we square them, we have that:

(6)
\begin{align} \quad \biggr | \int_{-1}^1 p(x) q(x) \: dx \biggr |^2 ≤ \left ( \int_{-1}^{1} (p(x))^2 \: dx \right ) \left (\int_{-1}^{1} (q(x))^2 \: dx \right ) \end{align}

## Example 2

Using the Cauchy-Schwarz inequality prove that $(x_1 + x_2 + ... + x_n)^2 ≤ n(x_1^2 + x_2^2 + ... + x_n^2)$.

Let $V$ be the vector space $\mathbb{R}^n$ of $n$ component vectors whose components are real numbers. Define an inner product on $\mathbb{R}^n$ by:

(7)
\begin{align} \quad <(x_1, x_2, ..., x_n), (y_1, y_2, ..., y_n)> = x_1y_1 + x_2y_2 + ... + x_ny_n \end{align}

This is just the inner product on $\mathbb{R}^n$. Then we have that:

(8)
\begin{align} \quad \| (x_1, x_2, ..., x_n \| = \sqrt{<(x_1, x_2, ..., x_n), (x_1, x_2, ..., x_n)>} = \sqrt{x_1^2 + x_2^2 + ... + x_n^2} \end{align}

Also we have that for the vector $(1, 1, ..., 1)$:

(9)
\begin{align} \quad \| (1, 1, ..., 1) \| = \sqrt{<(1, 1, ..., 1), (1, 1, ..., 1)>} = \sqrt{1^2 + 1^2 + ... + 1^2} = \sqrt{n} \end{align}

Therefore by the Cauchy-Schwarz Inequality we have that

(10)
\begin{align} \quad \mid <(x_1, x_2, ..., x_n), (1, 1, ..., 1)> ≤ \| (x_1, x_2, ..., x_n) \| \| (1, 1, ..., 1) \| \\ \quad (x_1 + x_2 + ... + x_n) ≤ \left ( \sqrt{x_1^2 + x_2^2 + ... + x_n^2} \right ) \left ( \sqrt{n} \right ) \\ \quad (x_1 + x_2 + ... + x_n)^2 ≤ n (x_1^2 + x_2^2 + ... +x_n^2) \end{align}

## Example 3

Prove that for all positive numbers $a, b, c, d \in \mathbb{R}$ that $16 ≤ (a + b + c + d) \left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right )$.

Let $V$ be the vector space $\mathbb{R}^4$, and define an inner product on $\mathbb{R}^4$ by:

(11)
\begin{align} \quad <(x_1, x_2, x_3, x_4), (y_1, y_2, y_3, y_4)> = x_1y_1 + x_2y_2 + x_3y_3 + x_4y_4 \end{align}

This is simply the dot product on $\mathbb{R}^2$. We therefore have that:

(12)
\begin{align} \| (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}) \| = \sqrt{<(\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}), (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d})>} = \sqrt{a + b + c + d} \end{align}
(13)
\begin{align} \quad \biggr \| \left ( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right ) \biggr \| = \sqrt{<\left ( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right ),\left ( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right )>} = \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \end{align}

Thus by the Cauchy-Schwarz inequality we have that:

(14)
\begin{align} \quad \quad \biggr | <(\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}), \left ( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right )> \biggr | ≤ \| (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}) \| \biggr \| \left ( \frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}} \right ) \biggr \| \\ \quad \quad \mid 1 + 1 + 1 + 1 \mid ≤ \left ( \sqrt{a+ b + c+ d} \right ) \left ( \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \right ) \\ \quad \quad \mid 4 \mid ≤ \left ( \sqrt{a+ b + c+ d} \right ) \left ( \sqrt{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}} \right ) \\ \quad \quad 16 ≤ (a + b + c + d) \left ( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} \right ) \end{align}