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Calculus/advanced calculus proof of paralellogram law
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Question
I need to show that:
||x -y||||x +y|| ≤ ||x||² +||y||²
can you show me how to do this? I'm not sure.
||x|| is the length of the vector x.
Thank You.
Hello, I moved your question to the question pool, but I sow that no
one solved it, so I decided to do that by my self.
Apologies for the lateness.
|x-y||x+y|≤|x+y||x+y|=|x+y|^2 = <x+y,x+y> = |x|^2+2<x,y>+|y|^2
So, |x-y||x+y|≤|x|^2+2<x,y>+|y|^2. From the Coachi-Schzartz inequality , we know that |<u,v>|<=|u||v|. Hence,
|x+y|^2≤|x|^2+2|x||y|+|y|^2≤|x|^2+|y|^2. Now we can claim that :
|x-y||x+y|≤|x+y||x+y|≤|x|^2+|y|^2 --> |x-y||x+y|≤|x|^2+|y|^2.
Q.E.D
Notes :
<x,y> represent an inner product.
|u| represent the Euclidean norm
Alon.
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Answers by Expert:
Alon Mandes
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