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I can't figure out why this proof of the CS inequality starts out as shown below. I understand everything afterwards. If you expand it you get an expression that is the proof. Can someone enlighten me?

$$\sum_{i=1}^n \sum_{j=1}^n\left(a_ib_j - a_jb_i\right)^2$$

Jeff Johnson
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    I don't know. Perhaps somebody can suggest an intuitively appealing reason for this. But you have to be prepared for the fact that sometimes, the only answer to "why does the proof do this?" is "because someone, somewhere, sometime was very smart". – David Dec 01 '22 at 23:29
  • The question isn't "why does the proof do this," it is why does it start out this way? Expanding it gives the right answer but why does this double sum represent the CS inequality? I've seen other proofs that make sense. – Jeff Johnson Dec 02 '22 at 02:05
  • OK, change "why does the proof do this?" to "why does the proof do this at the start?" if that makes it more comprehensible to you. Then that is exactly what your question is asking. – David Dec 02 '22 at 05:03

2 Answers2

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This might be sort of intuitive: consider the projections $ p_{ij} \colon \mathbb R^n \to \mathbb R^2 $, $ (x_1, \dotsc, x_n) \mapsto (x_i, x_j) $. Your sum vanishes iff $ \forall i, j \colon p_{i,j}(a_1, \dotsc, a_n) \parallel p_{i,j}(b_1, \dotsc, b_n) $. That happens iff $ (a_1, \dotsc, a_n) \parallel (b_1, \dotsc, b_n) $, which is precisely when the equality in CS holds. The "less parallel" the vectors are, the further the inequality is from equality.

nasekatnasushi
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  1. Continuing Nasekatnaushi's interesting idea with my interpretation, "less parallel" means, for example, perpendicular. Now, since $\langle b_i,b_j\rangle\cdot\langle b_j,-b_i\rangle =0$, it is natural to consider the procucts $\langle a_i,a_j\rangle\cdot\langle b_j,-b_i\rangle =a_ib_j-a_jb_i$ which are not in general zero and to consider the sum of their squares for the proof of CS inequality.
  2. In CS inequality, if we omit all terms not involving indicies $i$ and $j$, only $a_i^2b_j^2+a_j^2b_i^2\geq 2a_ia_jb_ib_j$ expression remains. This may give an idea for the start of the proof too.
Bob Dobbs
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