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I am studying cauchy inequality proof from notes I have from my class$$(\forall\vec{x},\vec{y}\in\mathbb{R}^n):|\sum_{i=1}^{n}x_iy_i|\le||\vec{x}||\cdot||\vec{y}||$$

We choose $\vec{x},\vec{y}$. And also this notation is used $<\vec{a},\vec{b}>=\sum_{i=1}^{n}a_ib_i$. For any $r\in\mathbb{R}$ we have: $$0\le <\vec{x}+r\vec{y},\vec{x}+r\vec{y}>=...=<\vec{y},\vec{y}>r^2+2<\vec{x},\vec{y}>r+<\vec{x},\vec{x}>$$ Now we set the discriminant $D\le0$ $$4<\vec{x},\vec{y}>-4<\vec{y},\vec{y}><\vec{x},\vec{x}>\le0$$ Here is one thing I don't undestand, since $D=b^2-4ac$ and $b=2<\vec{x},\vec{y}>$ why it's not $$4<\vec{x},\vec{y}>^2-4<\vec{y},\vec{y}><\vec{x},\vec{x}>\le0$$ It might be trivial but I don't see it(or I might have copied my notes wrong). The rest of the proof follows as this: $$<\vec{x},\vec{y}>\le<\vec{y},\vec{y}><\vec{x},\vec{x}>$$ $$\sum_{i=1}^{n}x_iy_i\le\sum_{i=1}^{n}x_i^2\sum_{i=1}^{n}y_i^2$$ $$\sum_{i=1}^{n}x_iy_i\le||\vec{x}||\cdot||\vec{y}||$$

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No, you're entirely right, and you probably just copied it down incorrectly. Here's the one point you're currently missing:

$$<x, x> = ||x||^2,$$

so if it weren't the square as you say it should be, the proof wouldn't work. Nice attention to detail typing all this up just to ask. :)

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  • $\begingroup$ Thank you,then I would have $$<\vec{x},\vec{y}>^2\le<\vec{y},\vec{y}><\vec{x},\vec{x}>$$ $$(\sum_{i=1}^{n}x_iy_i)^2\le\sum_{i=1}^{n}x_i^2\sum_{i=1}^{n}y_i^2$$ $$(\sum_{i=1}^{n}x_iy_i)^2\le||\vec{x}||\cdot||\vec{y}||$$, which isn't what I wanted to prove, or am I misunderstanding something about your response? $\endgroup$ – cgnx Jun 5 '14 at 22:01
  • $\begingroup$ @D.N In your first line, add on the RHS $= || x ||^2 || y ||^2$. If this doesn't instantly ring a bell, have another look at your definitions (the norm is the square root of the sum of squares). $\endgroup$ – gnometorule Jun 5 '14 at 22:20
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    $\begingroup$ I got it :), maybe I should think more and write less, thank you for your patience $\endgroup$ – cgnx Jun 5 '14 at 22:25

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