One commonly used proof in the discrete case is via Lagrange’s identity. Here are two other nice proofs.
1. For any complex number we use the non-negativity of the square :
Let . The above equation shows that if the 2 by 1 vector has non-zero first component, then . This non-negativity also holds when has zero first component as for we have . We conclude that A is positive semidefinite, so has non-negative determinant
from which (1) follows.
2. For convenience let . Consider the projection of onto , given by . We compute the squared length of their difference.
from which (1) easily follows. Equality holds iff the projection of onto is the zero vector: in other words, they are parallel or one of the vectors is zero.
Proof 2 shows that the Cauchy-Schwarz inequality is a consequence of Gram-Schmidt orthogonalisation applied to two vectors, also mentioned in .
 J. M. Steele, The Cauchy-Schwarz Master Class: An Introduction to the Art of Mathematical Inequalities, MAA Problem Books Series, 2004.