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?
Best Answer
Start with the algebraic identity (also known as the Brahmagupta-Fibonacci Identity) $$(x_1^2 + x_2^2)(y_1^2 + y_2^2) = (x_1y_1 + x_2y_2)^2 + (x_1y_2 - x_2y_1)^2$$
Then take away the $(x_1y_2 - x_2y_1)^2$ term as you did (here you use the fact that $a^2\ge0$ for all $a$) to give the inequality: $$(x_1y_1 + x_2y_2)^2\le(x_1^2 + x_2^2)(y_1^2 + y_2^2)$$
Then take the square root of both sides, since both are positives numbers. This next step of taking the modulus $|x_1y_1 + x_2y_2|$ is unecessary. You already have the Schwarz inequality here. Instead of
$$|x_1y_1 + x_2y_2|\le\sqrt{x_1^2 + x_2^2}\cdot\sqrt{y_1^2 + y_2^2}$$ you only need $$x_1y_1 + x_2y_2\le\sqrt{x_1^2 + x_2^2}\cdot\sqrt{y_1^2 + y_2^2}$$ as it doesn't matter whether the $x_i$ or $y_i$ are positive or negative, the Schwarz inequality still holds. Hence the $-b \le a \le b$ is then not needed, only the fact that $a^2\ge0$ for all $a$ is ever used.
The general form of the Schwarz Inequality is then: $$\sum_{i=1}^nx_iy_i\le\sqrt{\sum_{i=1}^nx_i^2}\cdot\sqrt{\sum_{i=1}^ny_i^2}$$