Cauchy-Schwarz Formula

One of the most important inequalities in mathematics is inarguably the famous Cauchy-Schwarz inequality whose use appears in many important proofs. We will prove this important inequality and prove an analogue of the triangle inequality in higher dimension Euclidean $n$-space.

The Cauchy-Schwarz Inequality

Theorem 1 (The Cauchy-Schwarz Inequality): If $\mathbf {x} =(x_{1},x_{2},...,x_{n}),\mathbf {y} =(y_{1},y_{2},...,y_{n})\in \mathbb {R} ^{n}$ then $(\mathbf {x} \cdot \mathbf {y} )^{2}\leq \|\mathbf {x} \|^{2}\|\mathbf {y} \|^{2}$.

• Proof: Let $\mathbf {x} ,\mathbf {y} \in \mathbb {R} ^{n}$. Then we want to prove that:

${\begin{aligned}\quad (\mathbf {x} \cdot \mathbf {y} )^{2}=\left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)=\|\mathbf {x} \|^{2}\|\mathbf {y} \|^{2}\quad (*)\end{aligned}}$

• Notice that the sum of squares is always nonnegative, and so for all $t\in \mathbb {R}$ we have that:

${\begin{aligned}\quad 0\leq \sum _{i=1}^{n}(x_{i}t+y_{i})^{2}=\sum _{i=1}^{n}(x_{i}^{2}t^{2}+2x_{i}y_{i}t+y_{i}^{2})=\sum _{i=1}^{n}x_{i}^{2}t^{2}+\sum _{i=1}^{n}2x_{i}y_{i}t+\sum _{i=1}^{n}y_{i}^{2}\end{aligned}}$

• Let $\displaystyle {A=\sum _{i=1}^{n}x_{i}^{2}}$, $\displaystyle {B=\sum _{i=1}^{n}x_{i}y_{i}}$, and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \displaystyle{C = \sum_{i=1}^{n} y_i^2}} . Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad 0 \leq At^2 + 2Bt + C \end{align}}

• Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A = 0} . Then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (*)} reduces to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 \leq 0} which is true. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A > 0} , then let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t = -\frac{B}{A}} . Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad 0 \leq A \left ( \frac{-B}{A} \right)^2 + 2B \left(\frac{-B}{A} \right ) + C = \frac{B^2}{A} - \frac{2B^2}{A} + C = - \frac{B^2}{A} + C \end{align}}

• Therefore we have that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad 0 \leq -\frac{B^2}{A} + C \\ \quad 0 \leq -B^2 + AC \\ \quad B^2 \leq AC \\ \quad \left ( \sum_{i=1}^{n} x_iy_i \right )^2 \leq \left ( \sum_{i=1}^{n} x_i^2 \right ) \left ( \sum_{i=1}^{n} y_i^2 \right ) \\ \quad (\mathbf{x} \cdot \mathbf{y})^2 \leq \| \mathbf{x} \|^2 \| \mathbf{y} \|^2 \quad \blacksquare \end{align}}

Often times the Cauchy-Schwarz inequality is stated by squaring both sides of the inequality above:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad \mid \mathbf{x} \cdot \mathbf{y} | \leq \| \mathbf{x} \| \| \mathbf{y} \| \end{align}}

The Triangle Inequality

Theorem 2 (The Triangle Inequality): If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{x} = (x_1, x_2, ..., x_n), \mathbf{y} = (y_1, y_2, ..., y_n) \in \mathbb{R}^n} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \| \mathbf{x} + \mathbf{y} \| \leq \| \mathbf{x} \| + \| \mathbf{y} \|} .

• Proof: Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{x}, \mathbf{y} \in \mathbb{R}^n} . Then:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad \| \mathbf{x} + \mathbf{y} \| = \| (x_1 + y_1, x_2 + y_2, ..., x_n + y_n) \| = \left ( \sum_{i=1}^{n} (x_i + y_i)^2 \right )^{1/2} \end{align}}

• Square both sides of the equation and apply the Cauchy-Schwarz inequality at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (*)} to get:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \quad \| \mathbf{x} + \mathbf{y} \|^2 = \sum_{i=1}^{n} (x_i + y_i)^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = (x_1 + y_1)^2 + (x_2 + y_2)^2 + ... + (x_n + y_n)^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = (x_1^2 + 2x_1y_1 + y_1^2) + (x_2^2 + 2x_2y_2 + y_2^2) + ... + (x_n^2 + 2x_ny_n + y_n^2) \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = (x_1^2 + x_2^2 + ... + x_n^2) + 2(x_1y_1 + x_2y_2 + ... + x_ny_n) + (y_1^2 + y_2^2 + ... + y_n^2) \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 = \| \mathbf{x} \|^2 + 2 (\mathbf{x} \cdot \mathbf{y}) + \| \mathbf{y} \|^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 \overset{(*)} \leq \| \mathbf{x} \|^2 + 2 \| \mathbf{x} \| \| \mathbf{y} \| + \| \mathbf{y} \|^2 \\ \quad \| \mathbf{x} + \mathbf{y} \|^2 \leq (\| \mathbf{x} \| + \| \mathbf{y} \|)^2 \end{align}}

• Square rooting both sides of the inequality above yields Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \| \mathbf{x} + \mathbf{y} \| \leq \| \mathbf{x} \| + \| \mathbf{y} \|} as desired. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Licensing

Content obtained and/or adapted from:

• The Cauchy-Schwarz and Triangle Inequalities, mathonline.wikidot.com under a CC BY-SA license