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In answers to various inequality questions, the solver often say "By C-S, ...". However, the inequality used as C-S is not always the same.

I have encountered at least three forms that people have called Cauchy-Schwarz. Let me denote the following as the "basic" form of C-S: $$ \left(\sum a_i^2 \right) \left(\sum x_i^2 \right) \geq \left(\sum a_i x_i \right)^2 $$ This is the form that appears on the wikipedia page, and in all the other links to Cauchy-Schwarz I can easily find.

However, often people state as C-S the inequality that for all $x_i > 0$, $$ \sum\left( \frac{a_i^2}{x_i}\right) \geq \frac{\left(\sum a_i \right)^2}{\sum x_i} $$ I can (see below) prove this directly, but since people call this C-S, there is probably a much simpler proof that starts from the basic form of C-S. What I would like know is:

Using the $ \left(\sum a_i^2 \right) \left(\sum x_i^2 \right) \geq \left(\sum a_i x_i \right)^2 $ basic C-S inequality, show a simple proof that $$ \sum\left( \frac{a_i^2}{x_i}\right) \geq \frac{\left(\sum a_i \right)^2}{\sum x_i} $$

Also

What other discrete inequalities are referred to as C-S, and again, how are they shown using the basic C-S?


To prove $ \sum\left( \frac{a_i^2}{x_i}\right) \geq \frac{\left(\sum a_i \right)^2}{\sum x_i} $ the "hard way," let $X \equiv \prod x_i$. Then: $$ \sum_{k<m} \frac{X}{x_kx_m}(a_kx_m-a_mx_k)^2 \\ \sum_{k<m}\left( \frac{Xx_m}{x_k}a_k^2 + \frac{Xx_k}{x_m}a_m^2 -2Xa_ka_m \right)\geq 0\\ \sum_i \left[ a_i^2\left( \sum_{j\neq i} \frac{Xx_j}{x_i} \right)\right] \geq \sum_{k<m}2Xa_ka_m \\ \sum_i \left[ a_i^2\left( \sum_{j\neq i} \frac{Xx_j}{x_i} \right)\right] +X\sum_i a_i^2 \geq X\sum_i a_i^2 +\sum_{k<m}2Xa_ka_m \\ \sum_i \left[ a_i^2\left( \sum_{j\neq i} \frac{Xx_j}{x_i} \right)\right] +X\sum_i a_i^2\frac{x_i}{x_i} \geq X\left( \sum_i a_i\right)^2 \\ X\sum_i \left[\frac{a_i^2}{x_i} \left( \sum_j x_j \right)\right] \geq X\left( \sum_i a_i\right)^2 \\ \sum\left( \frac{a_i^2}{x_i}\right) \geq \frac{\left(\sum a_i \right)^2}{\sum x_i} $$

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    $\begingroup$ Hint: $a_i=\sqrt{x_i}\frac{a_i}{\sqrt{x_i}}$ will take basic form to the desired form. Knowing $x_i>0$ is crucial. $\endgroup$ – Anurag A Jun 9 '17 at 21:15
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Let $y_i=\sqrt{x_i}$, and $b_i=\frac{a_i}{y_i}$. Then:

$$\left(\sum y_i^2\right)\left(\sum b_i^2\right)\geq\left(\sum y_ib_i\right)^2$$

is your inequality.

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  • $\begingroup$ Both answers are good; yours apparently was first. $\endgroup$ – Mark Fischler Jun 9 '17 at 21:47
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$$\left(\sum_i a_i\right)^2 = \left(\sum_i \frac{a_i}{\sqrt{x_i}} \sqrt{x_i}\right)^2 \overset{\text{C-S}}{\le} \left(\sum_i \frac{a_i^2}{x_i}\right) \left(\sum_i x_i\right).$$

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