Vtyjkyuk

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?

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 $a^2ge0$ for all $a$) to give the inequality:
$$(x_1y_1 + x_2y_2)^2le(x_1^2 + x_2^2)(y_1^2 + y_2^2)$$

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

$$|x_1y_1 + x_2y_2|lesqrtx_1^2 + x_2^2cdotsqrty_1^2 + y_2^2$$
you only need
$$x_1y_1 + x_2y_2lesqrtx_1^2 + x_2^2cdotsqrty_1^2 + y_2^2$$
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 $a^2ge0$ for all $a$ is ever used.

The general form of the Schwarz Inequality is then:
$$sum_i=1^nx_iy_ilesqrtsum_i=1^nx_i^2cdotsqrtsum_i=1^ny_i^2$$

$(x_1,x_2)cdot (y_1,y_2)=mid xmidcdotmid ymidcosthetaimplies mid x_1y_1+x_2y_2midle sqrt(x_1^2+x_2^2)(y_1^2+y_2^2)$

Is that ok?