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I understand common proofs for Cauchy-Schwarz, but not sure about the first step in this one, which is proof 4 from here

Let $A = \sqrt{a_1^2 + a_2^2 + \dots + a_n^2}$ and $B = \sqrt{b_1^2 + b_2^2 + \dots + b_n^2}$. By the arithmetic-geometric means inequality (AGI), we have

$$ \sum_{i=1}^n \frac{a_ib_i}{AB} \leq \sum_{i=1}^n \frac{1}{2} \left( \frac{a_i^2}{A^2} + \frac{b_i^2}{B^2} \right) = 1 $$

so that

$$ \sum_{i=1}^na_ib_i \leq AB \leq \sqrt{\sum_{i=1}^na_i^2} \sqrt{\sum_{i=1}^n b_i^2} $$

Which is Cauchy-Schwarz. Now this is all quite elegant, but how is the first equation RHS equal to 1? And how is it AGI? Shouldn't it be in this form:

$$ \sqrt[n]{\prod_{i=1}^n x_i} \leq \frac{1}{n} \sum_{i=1}^n x_i $$

There's probably something simple I'm missing...

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First, $$\sum_{i=1}^{n} \frac{a_i^2}{A^2} = \frac{1}{A^2}\sum_{i=1}^{n} a_i^2 = \frac{A^2}{A^2} = 1.$$ Similar could be said about the second term in the summation.

Second, the proof uses the following two-variable AM-GM inequality: $$xy = \sqrt{x^2 y^2} \le \frac{x^2+y^2}{2}.$$

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  • $\begingroup$ except there is also the trivial inequality $x\leq |x|$ since $a_i,b_i$ are not assumed to be nonegative. $\endgroup$ Sep 21, 2018 at 18:46
  • $\begingroup$ thanks, but what would x be? $\endgroup$
    – vega
    Sep 21, 2018 at 20:28
  • $\begingroup$ @vega $x=\frac{a_i}{A}$ and $y = \frac{b_i}{B}$. $\endgroup$
    – Math Lover
    Sep 21, 2018 at 21:34
  • $\begingroup$ I don't fully get how we assume this extends to sums, but ok $\endgroup$
    – vega
    Sep 21, 2018 at 22:02

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