Spivak Calculus (World Student Series, 1e, 1967): Part I, Chapter 1, Problem 18 c):
Prove the Schwartz inequality by using $2xy \le x^2 + y^2$ (how is this derived?) with
$$ x = \frac{x_i}{\sqrt{x_1^2 + x_2^2}}, \qquad y = \frac{y_i}{\sqrt{y_1^2 + y_2^2}}, $$ first for $i = 1$ and then for $i = 2$.
Here is the solution taken from the answer book.
I gave it myself a try and came up with the following lengthier version:
Assume the Schwarz inequality not to hold:
$$ (\exists x_1, x_2, y_1, y_2 \in \mathbb{R}) \; 0 \le \sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)} < x_1y_1 + x_2y_2. $$
Then
$$ (x_1^2 + x_2^2)(y_1^2 + y_2^2) < (x_1y_1 + x_2y_2)^2. $$
After some reformulations:
$$ x_1^2(y_1^2 + y_2^2) + y_1^2(x_1^2 + x_2^2) < 2x_1y_1(x_1y_1 + x_2y_2) $$
Introducing
$x := \frac{x_1}{\sqrt{x_1^2 + x_2^2}}, \qquad y := \frac{y_1}{\sqrt{y_1^2 + y_2^2}}$,
I'm ending up with
$$ x^2 + y^2 < 2xy \underbrace{\frac{x_1y_1 + x_2y_2}{\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)}}}_{>1}. $$
Since this inequality holds, e.g. for $x = y = 1$, the assumption holds for certain values of $x_1, x_2, y_1, y_2$.
Which causes the proof-by-contradiction to fail. Can you see why?