Cauchy-Schwarz Inequality Proof
Theorem
$$ ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})\ge { (ax+by) }^{ 2 } $$
Proof
$$ \begin{align*} & ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})-{ (ax+by) }^{ 2 } \\ =& {a}^{2}{x}^{2}+{b}^{2}{x}^{2}+{a}^{2}{y}^{2}+{b}^{2}{y}^{2}-{ (ax+by) }^{ 2 } \\ =& {b}^{2}{x}^{2}+{a}^{2}{y}^{2}-2axby \\ =& { (ay-bx) }^{ 2 } \\ \ge& 0 \end{align*} $$ Thus, we can summarize as follows. $$ ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})\ge { (ax+by) }^{ 2 } $$
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Explanation
This inequality, which can be encountered as early as high school, is used widely across various fields. Its algebraic proof is very straightforward.
As can be seen in the proof process, equality holds only in the case of $ay-bx=0$. The Cauchy-Schwarz inequality can also be expressed in the form of an equation, including the terms that appear in the proof.
$$ ({a}^{2}+{b}^{2})({x}^{2}+{y}^{2})={ (ax+by) }^{ 2 }+{ (ay-bx) }^{ 2 } $$
This implies that a sum of some squares can be represented as the product of sums of other squares.