Teacher proved it like this:

enter image description here

Very elegant (way simpler than most of the ones I find online), but I'm still not convinced -- particularly the last two steps or so, where the absolute value on the left-hand side seems to disappear.

Any thoughts?

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    $\begingroup$ Hint: What's the absolute value of a nonnegative number? $\endgroup$ – Ennar Apr 2 at 17:36
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    $\begingroup$ It is so elegant that it is wrong. If $\cos \theta$ is negative, then $\frac{n}{\cos \theta}\ge n$ is false. Besides, the fact that you can write $$\vec{a}\cdot\vec{b}=|a||b|\cos \theta$$ relies on the Cauchy-Schwarz inequality, so this reasoning (if it were correct) would be circular. $\endgroup$ – Giuseppe Negro Apr 2 at 17:38
  • $\begingroup$ @Ennar: But how do we know the dot product of vectors a and b IS nonnegative? Dot products can sometimes be negative, after all. $\endgroup$ – Will Apr 2 at 17:38
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    $\begingroup$ @GiuseppeNegro: Let's be charitable and assume that there might have been a transcription error, and the actual claim is that $\frac{n}{| \cos \theta |} \ge n$ if $n \ge 0$. Then the remainder of the argument follows, although the reasoning is, as you point out, fundamentally circular. $\endgroup$ – John Hughes Apr 2 at 17:55
  • $\begingroup$ How did the teacher define $\vec a\cdot\vec b $, $\lVert\vec a\rVert $ and $\theta $? $\endgroup$ – user Apr 3 at 5:22

given that $a\cdot b= \|a\|\cdot \|b\|\cos(\theta)$ we have that $$|a\cdot b|= \|a\|\cdot \|b\||\cos(\theta)|$$ (since the length of a vector is always non-negative).

Noting that $|\cos(\theta)|\leq 1$ we can have that:

  1. $|\cos(\theta)|=1$ which implies that $|a\cdot b|\leq\|a\|\cdot \|b\|$ is satisfied with equality.
  2. $|\cos(\theta)|<1$ which implies that $|a\cdot b|<\|a\|\cdot \|b\|$

The latter proves the statement.

$$|a\cdot b| \leq \|a\|\cdot \|b\|$$

EDIT: As pointed out by @user in the comments down below, there's no sense in moving the $|\cos(\theta)|$ to the denominator as your teacher suggested, cause nothing prevents you to have $\theta=\frac{\pi}2$ implying that $\cos(\theta)=0$. It's better to keep it on the other side and deriving the conclusion from there.

Disclaimer I think you should,nevertheless, consider what @jose-carlos-santos and @giuseppe-negro are pointing out, that actually this is a circular reasoning, and hence not a valid proof.

  • $\begingroup$ Why does the fact that cosine is less than or equal to one imply the final statement? I don't quite follow. $\endgroup$ – Will Apr 2 at 17:46
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    $\begingroup$ I've edited the answer to be more clear $\endgroup$ – RScrlli Apr 2 at 17:51
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    $\begingroup$ Why should one divide both sides of the equality by $|\cos\theta|$ introducing possible singularity? This step is absolutely excessive even assuming that both expressions $\vec a\cdot\vec b=\lVert\vec a\rVert\,\lVert\vec b\rVert\cos\theta $ and $-1\le\cos\theta\le1$ are "given". $\endgroup$ – user Apr 3 at 5:16
  • $\begingroup$ I was just replicating the proof given in the sketch of his professor. $\endgroup$ – RScrlli Apr 3 at 5:55

This proof assumes that $\vec a.\vec b$ can be written as $\left\lVert\vec a\right\rVert.\left\lVert\vec b\right\rVert.\cos\theta$ for some number $\theta$. This is the same thing as asserting that $\left\lvert\vec a.\vec b\right\rvert\leqslant\left\lVert\vec a\right\rVert.\left\lVert\vec b\right\rVert$, and this is precisely the Cauchy-Schwarz inequality, which is what you want to prove. There is therefore a circular reasoning here.


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