# Cauchy-Schwarz Inequality

The Cauchy-Schwarz inequality is among the most useful inequalities in all of mathematics. Suppose $$V$$ is a (real) vector space and $$\langle \cdot, \cdot \rangle$$ be an inner product on $$V$$. That is, for all $$\mathbf{u}, \mathbf{v}, \mathbf{w} \in V$$ and $$a, b \in \mathbf{R}$$, the inner product $$\langle \cdot, \cdot \rangle : V \times V \to \mathbf{R}$$ satisfies the following properties:

1. $$\langle \mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{v} ,\mathbf{u} \rangle$$
2. $$\langle a \mathbf{u} + b \mathbf{v}, \mathbf{w} \rangle = a \langle \mathbf{u}, \mathbf{w} \rangle + b \langle \mathbf{v}, \mathbf{w} \rangle$$
3. $$\langle \mathbf{u}, \mathbf{u} \rangle \geq 0$$ with $$\langle \mathbf{u}, \mathbf{u} \rangle = 0$$ if and only if $$\mathbf{u} = \mathbf{0}$$

Under these conditions, the Cauchy-Schwarz inequality states that
$|\langle \mathbf{u}, \mathbf{v} \rangle| \leq |\mathbf{u}| |\mathbf{v}|.$
Here, the norm of a vector is given by $$|\mathbf{v}|^2 = \langle \mathbf{v}, \mathbf{v} \rangle$$.

To prove the inequality, we employ the following trick: For any $$\lambda \in \mathbf{R}$$, we have
$\langle \mathbf{u} + \lambda \mathbf{v}, \mathbf{u} + \lambda \mathbf{v} \rangle \geq 0$
by property 3. The Cauchy-Schwarz inequality follows from choosing a suitable value for $$\lambda$$. Using properties 1 and 2, we can expand
$\langle \mathbf{u} + \lambda \mathbf{v}, \mathbf{u} + \lambda \mathbf{v} \rangle = \langle \mathbf{u}, \mathbf{u} \rangle + 2 \lambda \langle \mathbf{u}, \mathbf{v} \rangle + \lambda^2 \langle \mathbf{v}, \mathbf{v} \rangle \geq 0.$
Notice that this is a quadratic equation in the variable $$\lambda$$. Recall that the quadratic equation $$a \lambda^2 + b \lambda + c$$ obtains its minimum (for $$a > 0$$) when $$\lambda = \frac{- b}{2 a}$$. Therefore, we take
$\lambda = – \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle}.$
Plugging this into the previous expression, we obtain
$\langle \mathbf{u}, \mathbf{u} \rangle – 2 \frac{\langle \mathbf{u}, \mathbf{v} \rangle}{\langle \mathbf{v}, \mathbf{v} \rangle} \langle \mathbf{u}, \mathbf{v} \rangle + \frac{\langle \mathbf{u}, \mathbf{v} \rangle^2}{\langle \mathbf{v}, \mathbf{v} \rangle^2} \langle \mathbf{v}, \mathbf{v} \rangle \geq 0.$
Multiplying both sides by $$\langle \mathbf{v}, \mathbf{v}\rangle$$ and rearranging gives
$\langle \mathbf{u}, \mathbf{v} \rangle^2 \leq \langle \mathbf{v}, \mathbf{v} \rangle \langle \mathbf{u}, \mathbf{u} \rangle$
whence the Cauchy-Schwarz inequality immediately follows.