Proof of the Cauchy-Schwarz Inequality
Theorem
$$ ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})\ge { (ax+by) }^{ 2 } $$
Proof
$$ \begin{align*} & ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})-{ (ax+by) }^{ 2 } \\ =& {a}^{2}{x}^{2}+{b}^{2}{x}^{2}+{a}^{2}{y}^{2}+{b}^{2}{y}^{2}-{ (ax+by) }^{ 2 } \\ =& {b}^{2}{x}^{2}+{a}^{2}{y}^{2}-2axby \\ =& { (ay-bx) }^{ 2 } \\ \ge& 0 \end{align*} $$ Rearranging this yields the following. $$ ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})\ge { (ax+by) }^{ 2 } $$
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Explanation
This is an inequality one encounters as early as high school, and it is used in many places regardless of field. The algebraic proof is very simple.
As can be seen from the proof, equality holds only when $ay-bx=0$. The Cauchy-Schwarz inequality can also be written in the form of an equation by including the term that appears in the proof.
$$ ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})={ (ax+by) }^{ 2 }+{ (ay-bx) }^{ 2 } $$
This implies that a sum of squares can be expressed as a product of sums of squares.