The inequality

(\sum_{i=1}^{n}{a_{i}b_{i}})^2 \le (\sum_{i=1}^{n}{a_i}^{2})(\sum_{i=1}^{n}{b_i}^{2})

is a direct consequence of Lagrange Identity and is known as Cauchy's Inequality. The equality is only reached when the vectors a = (a_1, \ldots, a_n) and a = (b_1, \ldots, b_n) are proportional.

Here is a simple independent proof. Let

f(t) = \sum_{i=1}^{n}(ta_{i} + b_{i})^2.


f(t) = t^2 \sum_{i=1}^{n}a_{i}^{2} + 2t \sum_{i=1}^{n} a_{i}b_{i} + \sum_{i=1}^{n}b_{i}^{2}.

f(t) is a quadratic polynomial and, by the definition f(t) \ge 0  for all real t,  meaning that its discriminant D = (\sum_{i=1}^{n} a_{i}b_{i})^2 - \sum_{i=1}^{n}a_{i}^{2}\sum_{i=1}^{n}b_{i}^{2}  is never positive: D \le 0.  But this is exactly Cauchy's inequality.