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 $\|\mathbf {x} +\mathbf {y} \|\leq \|\mathbf {x} \|+\|\mathbf {y} \|$.

• Proof: Let $\mathbf {x} ,\mathbf {y} \in \mathbb {R} ^{n}$. Then:

${\begin{aligned}\quad \|\mathbf {x} +\mathbf {y} \|=\|(x_{1}+y_{1},x_{2}+y_{2},...,x_{n}+y_{n})\|=\left(\sum _{i=1}^{n}(x_{i}+y_{i})^{2}\right)^{1/2}\end{aligned}}$

• Square both sides of the equation and apply the Cauchy-Schwarz inequality at $(*)$ to get:

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

• Square rooting both sides of the inequality above yields $\|\mathbf {x} +\mathbf {y} \|\leq \|\mathbf {x} \|+\|\mathbf {y} \|$ as desired. $\blacksquare$

Licensing

Content obtained and/or adapted from:

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