Prove the Cauchy-Schwarz Inequality

Linear Algebra Problems and Solutions

Problem 355

Let $\mathbf{a}, \mathbf{b}$ be vectors in $\R^n$.
Prove the Cauchy-Schwarz inequality:
\[|\mathbf{a}\cdot \mathbf{b}|\leq ||\mathbf{a}||\,||\mathbf{b}||.\]

Here $\mathbf{a}\cdot \mathbf{b}$ is the dot (inner) product of $\mathbf{a}$ and $\mathbf{b}$, and $||\mathbf{a}||$ is the length (norm) of the vector $\mathbf{a}$.

 
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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})\\[6pt] &=\mathbf{a}\cdot \mathbf{a}-\mathbf{a}\cdot x\mathbf{b}-x\mathbf{a}\cdot \mathbf{b}+x^2\mathbf{b}\cdot \mathbf{b}\\[6pt] &=||\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\\[6pt] &=-\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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