While tricks have names because they wind up being associated with with some particular mathematician, tricks are tricks because something important goes on "behind the curtain."

For instance, to prove $$ (a_1 b_1 + \cdots + a_n b_n)^2 \leq ({a_1}^2 + \cdots + {a_n}^2) ({b_1}^2 + \cdots + {b_n}^2), $$ write \begin{align*} A &= ({a_1}^2 + \cdots + {a_n}^2)\\\ B &= (a_1 b_1 + \cdots + a_n b_n)\\\ C &= ({b_1}^2 + \cdots + {b_n}^2), \end{align*} then we must show $$ B^2 \leq AC. $$ Equality clearly holds when $A = 0$. Otherwise, since $\mathbb{R}$ has no negative squares, for all $x \in \mathbb{R}$, $$ 0 \leq (a_1 x - b_1)^2 + \cdots + (a_n x - b_n)^2. $$ Expanding the squares, $$ 0 \leq Ax^2 - 2Bx + C. $$

The quadratic expression vanishes whenever $$ x = \frac{B}{A} \pm > \sqrt{\left(\frac{B}{A}\right)^2 - > \frac{C}{A}}. $$

If $x = \dfrac{B}{A}$, then $$ 0 \leq A\left(\frac{B}{A}\right)^2 - 2 B\left(\frac{B}{A}\right) + C = \frac{B^2}{A} - 2 \frac{B^2}{A} + C = - \frac{B^2}{A} + C, $$ thus $$ B^2 \leq AC. $$