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How to prove $ |\langle u,v\rangle | \leq \|u\|\|v\|$

Note: I have given this many attempts so don't downvote due to lack of effort, refer to edit history for evidence of said effort

Mohsen Shahriari
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    The last second "identity" is wrong – Shuchang May 29 '14 at 06:01
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    For an extreme example, use $u=(1,0)$ and $v=(0,1)$. – André Nicolas May 29 '14 at 06:02
  • Why the 2 downvotes? I am trying to understand... –  May 29 '14 at 06:03
  • @AndréNicolas Very very good call. It seems I had forgotten something as simple as the purpose of magnitude! –  May 29 '14 at 06:08
  • Or more to the point, I was treating $u$ and $v$ as numbers and not vectors. –  May 29 '14 at 06:09
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    @user142198: your bigger mistake is not understanding the sigma operator and how square roots interact with sums. $(\sum x_i)(\sum y_i)\neq\sum x_iy_i$ and $\sqrt{\sum x_i}\neq\sum\sqrt{x_i}$. – symplectomorphic May 29 '14 at 06:11
  • You may want to look up the Cauchy-Schwarz Inequality on Wikipedia. – André Nicolas May 29 '14 at 06:11
  • Even if you were right about equality always holding, that wouldn't make the stated inequality "not true". – bof May 29 '14 at 06:24
  • @bof I know that, but the title should and final question should have evidenced what I was getting at. –  May 29 '14 at 06:29
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    I knew what you were "getting at". However, in math it's a good idea to say what you mean and mean what you say because math is hard, and us readers can easily get confused even if you say everything right, so there's no need to add to our confusion by saying things wrong. – bof May 29 '14 at 06:57
  • And where does $w$ come in? – Urgje May 29 '14 at 08:16
  • @bof say what I mean $\times 2$ because $p\rightarrow q =q\rightarrow p$ –  May 30 '14 at 01:15
  • @AndréNicolas I may misunderstand this myself, but don't $u,v$ have $n$ elements, and don't they need to be $(u_1^2,u_2^2,\dots,u_n^2)(v_1^2,v_2^2,\dots,v_n^2)$.

    Which equals $LHS: |<u,v>|$

    I am not sure, but you might be right user142198.

     – Display Name May 30 '14 at 01:58
  • Not with your steps I will note though.(As pointed out those steps are wrong.)

    Also I am almost certain you are making a joke, but I will note that $p \rightarrow q \not \equiv q \rightarrow p$

     – Display Name May 30 '14 at 02:07
  • The example I was giving was two-dimensional, and the inner product was the usual one. For general inner products, the inequality is called Cauchy-Schwarz, and there is a proof in the article I linked to. – André Nicolas May 30 '14 at 02:43
  • Duplicate. http://math.stackexchange.com/questions/436559/a-natural-proof-of-the-cauchy-schwarz-inequality – ClassicStyle May 30 '14 at 03:04
  • Does this answer your question? Proof Cauchy-Schwarz inequality –  Jul 25 '20 at 22:04

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Your $$\sqrt{\sum_{i=1}^n u_i u_i} \sqrt{\sum_{i=1}^n v_i v_i} = \sqrt{\sum_{i=1}^n u_i^2 v_i^2}$$ and $$ \sqrt{\sum_{i=1}^n u_i^2 v_i^2} = \sum_{i=1}^n u_i v_i$$ are both wrong.

Robert Israel
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  • On reflection they are both silly. Is that the only scenario that is wrong? –  May 29 '14 at 06:05