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.

Will Rosenbaum

Tel Aviv

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