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 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 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 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 \in \mathbb{R}} 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 \sum_{i=1}^{n} (x_it + y_i)^2 = \sum_{i=1}^{n} (x_i^2 t^2 + 2x_iy_it + y_i^2) = \sum_{i=1}^{n} x_i^2 t^2 + \sum_{i=1}^{n} 2x_iy_i t + \sum_{i=1}^{n} y_i^2 \end{align}}

• 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 \displaystyle{A = \sum_{i=1}^{n} x_i^2}} , 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{B = \sum_{i=1}^{n} x_iy_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