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.

$$\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)} \ge |x_1y_1 + x_2y_2| $$


$$\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2²)} \ge (x_1y_1 + x_2y_2) \ge -\sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)}$$

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?

  • $\begingroup$ Looks fine. Except, at the end, don't you mean $\mid a\mid\le b$. Also, not sure what you did after that. $\endgroup$ – Chris Custer Aug 30 '18 at 4:36
  • $\begingroup$ Oh, sorry, it's $|a| \le b$ $\endgroup$ – Jonatan Garcia Aug 30 '18 at 4:38
  • $\begingroup$ Already solved, I wanted to explain what I did. Is that so rigorous? $\endgroup$ – Jonatan Garcia Aug 30 '18 at 4:40
  • $\begingroup$ I like it. Did Spivak suggest this? $\endgroup$ – Chris Custer Aug 30 '18 at 4:41
  • 1
    $\begingroup$ two points: 1) this method works only for binomials (e.g. it will not work for $(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)$; 2) you must state when the equality holds ($x_1y_2-x_2y_1=0$). Also, you can check this: rgmia.org/papers/v12e/Cauchy-Schwarzinequality.pdf $\endgroup$ – farruhota Aug 31 '18 at 17:57

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: $$\sum_{i=1}^nx_iy_i\le\sqrt{\sum_{i=1}^nx_i^2}\cdot\sqrt{\sum_{i=1}^ny_i^2}$$

  • $\begingroup$ Echoing @farruhota s comments-this method only works for binomials as given (I only added the general form for reference). $\endgroup$ – Daniel Buck Aug 31 '18 at 19:30

$(x_1,x_2)\cdot (y_1,y_2)=\mid x\mid\cdot\mid y\mid\cos\theta\implies \mid x_1y_1+x_2y_2\mid\le \sqrt{(x_1^2+x_2^2)(y_1^2+y_2^2)}$

Is that ok?

  • $\begingroup$ I don't get what did you do $\endgroup$ – Jonatan Garcia Aug 30 '18 at 5:21
  • $\begingroup$ I used the formula for the dot product. Then I used that $\cos\theta\le1$. ($\theta$ is the angle between $x$ and $y$). It's an alternate proof (I think). $\endgroup$ – Chris Custer Aug 30 '18 at 5:25

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