Cauchy-Schwarz via sums of squares

Lagrange's identity.
$$(a^2+b^2)(c^2+d^2)-(ac+bd)^2=a^2c^2+a^2d^2+b^2c^2+b^2d^2-a^2c^2-2abcd-b^2d^2$$
$$=a^2d^2+b^2c^2-2abcd=(ad-bc)^2.$$

Inequality. Since $(ad-bc)^2\ge0$, we have
$$(a^2+b^2)(c^2+d^2)\ge(ac+bd)^2.$$

Equality case. $ad=bc$, i.e. $(a,b)$ and $(c,d)$ are proportional.

Generalization. In $n$ dimensions, Lagrange's identity is
$$\left(\sum a_i^2\right)\left(\sum b_i^2\right)-\left(\sum a_ib_i\right)^2=\sum_{i<j}(a_ib_j-a_jb_i)^2.$$