Applications of Cauchy-Schwarz: Means and Averages

In this post, we continue exploring the power of the Cauchy-Schwarz Inequality. While Part 1 focused on basic linear bounds, Part 2 dives into how this inequality governs the relationship between different types of averages.

Problems

1. Bounds for Reciprocal Sums

Suppose $a, b, c, d$ are positive numbers.
(a) Prove that $(a + b + c + d)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right) \geq 16$.
(b) For which positive numbers $a, b, c, d$ is the inequality above an equality?

2. Square of Average vs. Average of Squares

Show that if $a_1, \dots, a_n \in \mathbb{R}$, then the square of the average of $a_1, \dots, a_n$ is less than or equal to the average of $a_1^2, \dots, a_n^2$: \(\left( \frac{a_1 + \dots + a_n}{n} \right)^2 \leq \frac{a_1^2 + \dots + a_n^2}{n}\)

Solutions

1. Bounds for Reciprocal Sums

Part (a)

Define two vectors $x$ and $y$ such that their inner product leads to a constant value.

Let \(x = (\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{d}), \quad y = \left(\frac{1}{\sqrt{a}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}}, \frac{1}{\sqrt{d}}\right)\).

Using the Cauchy-Schwarz inequality \(|\langle x, y \rangle|^2 \leq \|x\|^2 \|y\|^2\),

\[\begin{aligned} |\langle x, y \rangle|^2 &\leq \|x\|^2 \|y\|^2 \\ \left| \left(\sqrt{a}\,\frac{1}{\sqrt{a}}\right) + \left(\sqrt{b}\,\frac{1}{\sqrt{b}}\right) + \left(\sqrt{c}\,\frac{1}{\sqrt{c}}\right) + \left(\sqrt{d}\,\frac{1}{\sqrt{d}}\right) \right|^{2} &\leq (a + b + c + d)(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}) \\ (1+1+1+1)^{2} &\leq (a + b + c + d)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right) \\ 16 &\leq (a + b + c + d)\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d}\right) \\ \end{aligned}\]

Part (b)

If $a = b = c = d = 1$, then \((1 + 1 + 1 + 1)(1 + 1 + 1 + 1) = 16\).
More generally, whenever $x$ and $y$ are linearly dependent, which occurs when $a = b = c = d$.
Let $a = b = c = d = z$ for some positive $z \in \mathbb{R}$.
Then \((z + z + z + z)(\frac{1}{z} + \frac{1}{z} + \frac{1}{z} + \frac{1}{z}) = (4z)(\frac{4}{z}) = 16\).

2. Square of Average vs. Average of Squares

This is a fundamental result in statistics, proving that the second moment is always greater than or equal to the square of the first moment, that is $\text{Var}(X) = E[X^2] - (E[X])^2 \geq 0$.

Let $u = (a_1, a_2, \dots, a_n)$ and let $v = (1, 1, \dots, 1) \in \mathbb{R}^n$.

Applying Cauchy-Schwarz,

\[\begin{aligned} |\langle x, y \rangle|^2 &\leq \|x\|^2 \|y\|^2 \\ (a_1(1) + a_2(1) + \dots + a_n(1))^2 &\leq (a_1^2 + a_2^2 + \dots + a_n^2)(1^2 + 1^2 + \dots + 1^2) \\ (a_1 + a_2 + \dots + a_n)^2 &\leq (a_1^2 + a_2^2 + \dots + a_n^2)(n) \\ \end{aligned}\]

Dividing both sides by $n^2$,

\[\begin{aligned} \frac{(a_1 + a_2 + \dots + a_n)^2}{n^2} &\leq \frac{n(a_1^2 + a_2^2 + \dots + a_n^2)}{n^2} \\ \left( \frac{a_1 + \dots + a_n}{n} \right)^2 &\leq \frac{a_1^2 + \dots + a_n^2}{n} \end{aligned}\]

Note: This result shows that the square of the sample mean is always bounded by the mean of the squares. In portfolio theory, this relationship is a precursor to understanding how volatility reduces the geometric growth rate of a fund.