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 $\displaystyle {C=\sum _{i=1}^{n}y_{i}^{2}}$. Then:

${\begin{aligned}\quad 0\leq At^{2}+2Bt+C\end{aligned}}$

• Suppose that $A=0$. Then $(*)$ reduces to $0\leq 0$ which is true. If $A>0$, then let $t=-{\frac {B}{A}}$. Then:

${\begin{aligned}\quad 0\leq A\left({\frac {-B}{A}}\right)^{2}+2B\left({\frac {-B}{A}}\right)+C={\frac {B^{2}}{A}}-{\frac {2B^{2}}{A}}+C=-{\frac {B^{2}}{A}}+C\end{aligned}}$

• Therefore we have that:

${\begin{aligned}\quad 0\leq -{\frac {B^{2}}{A}}+C\\\quad 0\leq -B^{2}+AC\\\quad B^{2}\leq AC\\\quad \left(\sum _{i=1}^{n}x_{i}y_{i}\right)^{2}\leq \left(\sum _{i=1}^{n}x_{i}^{2}\right)\left(\sum _{i=1}^{n}y_{i}^{2}\right)\\\quad (\mathbf {x} \cdot \mathbf {y} )^{2}\leq \|\mathbf {x} \|^{2}\|\mathbf {y} \|^{2}\quad \blacksquare \end{aligned}}$

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

${\begin{aligned}\quad \mid \mathbf {x} \cdot \mathbf {y} |\leq \|\mathbf {x} \|\|\mathbf {y} \|\end{aligned}}$

The Triangle Inequality

Theorem 2 (The Triangle Inequality): If $\mathbf {x} =(x_{1},x_{2},...,x_{n}),\mathbf {y} =(y_{1},y_{2},...,y_{n})\in \mathbb {R} ^{n}$ 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