Proof of the Cauchy-Schwarz inequality
From Mathematics
If 
and 
are vectors in 
, then 
[edit] Proof
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
which is equal to the original statement.
QED
[edit] Notes
- ↑ 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.
