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Inequalities (2 Viewers)

jmromeo

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Hi everybody!....Please I need help with this:

Prove that (x^2+y^2+z^2)(a^2+b^2+c^2)>=(ax+by+cz)^2

Thanks!
 

math man

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expand LHS, then group the non ax, by, cz terms and apply cauchy swcharz inequality on them

or you could even expand the RHS and apply cauchy schwartz inequality on the non squared terms then factorise
 
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Carrotsticks

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expand LHS, then group the non ax, by, cz terms and apply cauchy swcharz inequality on them
It actually IS the C.S Inequality in itself (I suppose you have to square both sides). But I was just kidding when I said to use it, obviously somebody trying to pull that off in the HSC would get 0 marks.

OP,

Consider the quadratic:



Note that it has either one real root, or no real roots. Hence, the discriminant is less than 0.

Expand the quadratic and group it so you have it in the form P(t) = At^2 + Bt + C for some value of A,B,C whatever it is, then let the discriminant be less than or equal to 0.

The required inequality falls out immediately.
 

math man

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i thought you were referring to the x^2 +y^2 >= 2xy simple cauchy schwartz inequality
 

math man

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it is a simple form if you put the vectors x and y in R2 and evaluate the dot product and norm
 

Fus Ro Dah

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it is a simple form if you put the vectors x and y in R2 and evaluate the dot product and norm
I don't think this is correct. The identity you gave simply comes from , and this is by no means the inequality you mentioned.
 
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seanieg89

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(I know it is trivial from expanding (x-y)^2, but it is also a consequence of two dimensional C-S by taking the dot product of (x,y) with (y,x). Perhaps that is what he meant.)
 

Fus Ro Dah

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(I know it is trivial from expanding (x-y)^2, but it is also a consequence of two dimensional C-S by taking the dot product of (x,y) with (y,x). Perhaps that is what he meant.)
That makes a lot more sense thanks for clearing that up.
 
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ORRR Take the RHS over to the LHS, Consider LHS-RHS, EXPAND!!! write nearly and clearly and don't miss terms. You'll end up with some perfect square. Ugly and inelegant but gets the job done. :party:
 

zhiying

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Just wondering if you use dot product in 4U do you get marks O_O
 

karnbmx

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ORRR Take the RHS over to the LHS, Consider LHS-RHS, EXPAND!!! write nearly and clearly and don't miss terms. You'll end up with some perfect square. Ugly and inelegant but gets the job done. :party:
I heard a LOT of people do that to get away from doing some very weird proofs from scratch. It even has a name (the reverse Snake method :p). However, do you really get marks for doing it in the HSC?

If you do, SHOULD you be awarded marks for doing that? because it isn't really exactly PROVING anything. You start from the result that you are given, so yeah...
 

Shadowdude

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I heard a LOT of people do that to get away from doing some very weird proofs from scratch. It even has a name (the reverse Snake method :p). However, do you really get marks for doing it in the HSC?

If you do, SHOULD you be awarded marks for doing that? because it isn't really exactly PROVING anything. You start from the result that you are given, so yeah...
Well you work backwards. Start with that horrible ugly square and then do algebra to it to get the result you want - just make sure you don't do any weird things, and it should generally be fine.
 

Carrotsticks

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Alternatively to be cheap, you could begin with the question, but with the inequality the wrong way around. Then obtain a contradiction.
 

seanieg89

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I heard a LOT of people do that to get away from doing some very weird proofs from scratch. It even has a name (the reverse Snake method :p). However, do you really get marks for doing it in the HSC?

If you do, SHOULD you be awarded marks for doing that? because it isn't really exactly PROVING anything. You start from the result that you are given, so yeah...

How is it not proving anything? Showing that LHS-RHS is non-negative or non-positive is precisely a proof of the claim the quesiton makes. You are not making any unjustified assumptions.
 

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