I have to prove the Schwarz inequality with that, then I want to know if what I did was so rigorous or not (And if it's good).
$$(x_1^2 + x_2^2)(y_1^2 + y_2^2) = (x_1y_1 + x_2y_2)^2 + (x_1y_2 - x_2y_1)^2$$
Then, to me it seems, that if I take away the $(x_1y_2 - x_2y_1)^2$ term
I get
$$(x_1^2 + x_2^2)(y_1^2 + y_2^2) \ge (x_1y_1 + x_2y_2)^2 $$
Then take the square root of both sides, since both are positives numbers.
$$\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)} \ge |x_1y_1 + x_2y_2| $$
Then
$$\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2²)} \ge (x_1y_1 + x_2y_2) \ge -\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)}$$
Since $|a| \le b$ then $-b \le a \le b$
Then I got the Schwarz inequality . I think in this too
If $a,b \ge 0$ and $a = b$, then $a - a \le b$
In this case, we have three terms, all positives. Then $a = b + c$ and then $a \ge b + c - c$
What do you thing about this?