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 July 8th, 2017, 04:12 AM #1 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,396 Thanks: 100 Cauchy Schwarz Inequality Cauchy Schwarz Inequality in Rn: $\displaystyle |x\cdot y|\leq |x||y|$ Proof: $\displaystyle \left | x\cdot \frac{y}{|y|}\right |\leq |x|$ Explanation: $\displaystyle \frac{y}{|y|}$ is a unit vector which can be expanded to a basis. Then either x is in the direction of y and equality holds, or it has a component among the rest of the basis in which case it's component along y is less than |x|.
 July 9th, 2017, 04:18 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 893 What was your purpose in posting this? Do you have a question about it?
September 26th, 2017, 09:30 PM   #3
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 Originally Posted by zylo Cauchy Schwarz Inequality in Rn: $\displaystyle |x\cdot y|\leq |x||y|$ Proof: $\displaystyle \left | x\cdot \frac{y}{|y|}\right |\leq |x|$ Explanation: $\displaystyle \frac{y}{|y|}$ is a unit vector which can be expanded to a basis. Then either x is in the direction of y and equality holds, or it has a component among the rest of the basis in which case it's component along y is less than |x|.
This is completely circular. Specifically "or it has a component among the rest of the basis in which case it's component along y is less than |x|" holds only if the Cauchy-Schwarz inequality is true.

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