Editing Cauchy–Schwarz inequality (section)

=== {{anchor|Real Euclidean space}}{{math|'''R'''<sup>''n''</sup>}}: ''n''-dimensional Euclidean space ===
{{anchor|real number proof}}In [[Euclidean space]] <math>\R^n</math> with the standard inner product, which is the [[dot product]], the Cauchy–Schwarz inequality becomes:
<math display=block>\biggl(\sum_{i=1}^n u_i v_i\biggr)^2 \leq \biggl(\sum_{i=1}^n u_i^2\biggr) \biggl(\sum_{i=1}^n v_i^2\biggr).</math>

The Cauchy–Schwarz inequality can be proved using only elementary algebra in this case by observing that the difference of the right and the left hand side is 
<math display=block> \tfrac{1}{2} \sum_{i=1}^n\sum_{j=1}^n (u_i v_j - u_j v_i)^2 \ge 0</math> 

or by considering the following [[quadratic polynomial]] in <math>x</math>
<math display=block> (u_1 x + v_1)^2 + \cdots + (u_n x + v_n)^2 = \biggl(\sum_i u_i^2\biggr) x^2 + 2 \biggl(\sum_i u_i v_i\biggr) x + \sum_i v_i^2.</math>

Since the latter polynomial is nonnegative, it has at most one real root, hence its [[discriminant]] is less than or equal to zero. That is,
<math display=block>\biggl(\sum_i u_i v_i\biggr)^2 - \biggl(\sum_i {u_i^2}\biggr)  \biggl(\sum_i {v_i^2}\biggr) \leq 0.</math>