# Prove the Cauchy-Schwarz Inequality

## Problem 355

Let $\mathbf{a}, \mathbf{b}$ be vectors in $\R^n$.

Prove the Cauchy-Schwarz inequality:
$|\mathbf{a}\cdot \mathbf{b}|\leq \|\mathbf{a}\|\,\|\mathbf{b}\|.$

Contents

We give two proofs.

## Proof 1

Let $x$ be a variable and consider the length of the vector $\mathbf{a}-x\mathbf{b}$ as follows.

We have
\begin{align*}
0& \leq \|\mathbf{a}-x\mathbf{b}\|^2=(\mathbf{a}-x\mathbf{b})\cdot (\mathbf{a}-x\mathbf{b})\\
&=\mathbf{a}\cdot \mathbf{a}-\mathbf{a}\cdot x\mathbf{b}-x\mathbf{a}\cdot \mathbf{b}+x^2\mathbf{b}\cdot \mathbf{b}\\
&=\|\mathbf{b}\|^2 x^2-2\mathbf{a}\cdot \mathbf{b}x+\|\mathbf{a}\|^2. \tag{*}
\end{align*}

Note that the last expression is an equation of a parabola (quadratic equation).
Since the parabola is always non-negative, its discriminant $D$ must be non-positive.

Hence we have
$D=(2\mathbf{a}\cdot \mathbf{b})^2-4\|\mathbf{a}\|^2 \|\mathbf{b}\|^2 \leq 0.$

It follows that we have
$(\mathbf{a}\cdot \mathbf{b})^2 \leq \|\mathbf{a}\|^2 \|\mathbf{b}\|^2 .$ Taking the square root, we obtain the Cauchy-Schwarz inequality
$|\mathbf{a}\cdot \mathbf{b}|\leq \|\mathbf{a}\|\,\|\mathbf{b}\|.$

## Proof 2

The second proof starts with the same argument as the first proof.
As in Proof 1 (*), we obtain
$0\leq \|\mathbf{b}\|^2 x^2-2\mathbf{a}\cdot \mathbf{b}x+\|\mathbf{a}\|^2.$

Now we take
$x=\frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{b}\|^2}.$ Then we have
\begin{align*}
0 &\leq \|\mathbf{b}\|^2 \left(\,\frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{b}\|^2} \,\right)^2-2\mathbf{a}\cdot \mathbf{b}\left(\, \frac{\mathbf{a}\cdot \mathbf{b}}{\|\mathbf{b}\|^2} \,\right)+\|\mathbf{a}\|^2\6pt] &=\frac{(\mathbf{a}\cdot \mathbf{b})^2}{\|\mathbf{b}\|^2}-2\frac{(\mathbf{a}\cdot \mathbf{b})^2}{\|\mathbf{b}\|^2}+\|\mathbf{a}\|^2\\ &=-\frac{(\mathbf{a}\cdot \mathbf{b})^2}{\|\mathbf{b}\|^2}+\|\mathbf{a}\|^2. \end{align*} It follows that we have \[(\mathbf{a}\cdot \mathbf{b})^2\leq \|\mathbf{a}\|^2 \|\mathbf{b}\|^2. The Cauchy-Schwarz inequality is obtained by taking the square root as in Proof 1.

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