# Proof of the Cauchy-Schwarz inequality

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If and are vectors in , then

## ProofEdit

If either or are the zero vector, the statement holds trivially, so assume that both and are nonzero. Let be a scalar and be defined as . Since, for any nonzero vector , (**NOTE**: merits own proof)

where , and . It can be seen clearly that is a quadratic polynomial that is non-negative for any . Consequently, the polynomial has two complex roots, or has a single distinct real root.^{[1]}

Remember that the roots of are given by the quadratic formula

In particular, the term must either be negative, yielding two complex roots, or zero, yielding a single real root. Thus

Substituting the values of , and into the last of these inequalities, it can be seen that

**Failed to parse (unknown function\lvert): \lvert \vec{X} \cdot \vec{Y} \rvert \leq \lVert \vec{X} \rVert \lVert \vec{Y} \rVert**

which is equal to the original statement.

QED

## NotesEdit

- ↑ Intuitively, the graph of is either 'floating above' the horizontal axis, if it has two complex roots, or tangent if it has one real root. Since is non-negative for every , it can't have two real roots because the graph of the function would have to 'pass under' the horizontal axis.