Cauchy-Bunyakovsky-Schwarz Inequality/Inner Product Spaces/Proof 2

Theorem
Let $\mathbb K$ be a subfield of $\C$.

Let $V$ be a semi-inner product space over $\mathbb K$.

Let $x, y$ be vectors in $V$.

Then:
 * $\left\vert{\left \langle {x, y} \right \rangle}\right\vert^2 \le \left \langle {x, x} \right \rangle \left \langle {y, y} \right \rangle$

Proof
This proof assumes that $V$ is a semi-inner product space over $\R$.

Then for all $x, y \in V$, we have $\left\langle{x, y}\right\rangle = \left\langle{y, x}\right\rangle$ by property $(1')$ of semi-inner products.

Define $f_{x, y}: \R \to \R_{\ge 0}$ by:


 * $f_{x, y} \left({\lambda}\right) = \left \langle{x - \lambda y, x - \lambda y}\right \rangle$

Then by property $(4)$ of semi-inner product:
 * $\forall \lambda \in \R: f_{x, y} \left({\lambda}\right) \ge 0$

For all $\lambda \in \R$, it follows that:

where we have put $a = \left \langle{y, y}\right \rangle$, $b = -2 \left\langle{x, y}\right\rangle$, and $c = \left \langle{x, x}\right \rangle$.

Then $f_{x,y}$ is a quadratic polynomial which satisfies $f_{x,y} \left({\lambda}\right) \ge 0$, so $f_{x,y}$ has at most one distinct real root.

From Solution to Quadratic_Equation, it follows that the discriminant $\Delta$ satisfies:


 * $\Delta = b^2 - 4ac \le 0$

Therefore:


 * $4 \left\langle{ x, y }\right\rangle ^ 2 - 4 \left \langle{ x, x }\right \rangle \left \langle{ y, y }\right \rangle \le 0$

which we rearrange as:


 * $\left\vert{\left\langle{x, y}\right\rangle }\right\vert^2 = \left\langle{x, y}\right\rangle^2 \le \left \langle{x, x}\right \rangle \left \langle{y, y}\right \rangle$