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    Got some questions and while I've done problem 1 and the first part of problem 2, I haven't been able to prove the Cauchy-Schwarz inequality nor could I proceed with the other problems.

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    So, starting from Q2, I'm not sure how I can go about it using the hint. I've tried doing v \cdot v which gave me

    \displaystyle (x-(x\cdot y)\lvert \lvert y \lvert \lvert ^{-2} \cdot y ) \cdot (x-(x\cdot y)\lvert \lvert y \lvert \lvert ^{-2} \cdot y)

    = (x\cdot x) -2(x \cdot y) \lvert \lvert y \lvert \lvert ^{-2}(y\cdot y)+ (\frac{x \cdot y}{\lvert \lvert y \lvert \lvert ^{2}})^2 (y \cdot y)\geq 0

    but I didn't seem to get anywhere so perhaps my approach is incorrect. Any guidance is appreciated, thanks.
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    (Original post by RDKGames)
    ..
    I think the fact that {\bf y} \cdot  {\bf y} = \vert\vert {\bf y} \vert \vert^2 will help you a lot here.
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    (Original post by DFranklin)
    I think the fact that {\bf y} \cdot  {\bf y} = \vert\vert {\bf y} \vert \vert^2 will help you a lot here.
    I have used it initially and got the inequality down to \vert\vert {\bf x} \vert \vert^2 - 2({\bf x}\cdot {\bf y})+({\bf x}\cdot {\bf y})^2 \vert\vert {\bf y} \vert\vert^{-2} \geq 0

    Does this seem right? I'm not sure how to manipulate it into the triangle inequality past this point.
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    (Original post by RDKGames)
    I have used it initially and got the inequality down to \vert\vert {\bf x} \vert \vert^2 - 2({\bf x}\cdot {\bf y})+({\bf x}\cdot {\bf y})^2 \vert\vert {\bf y} \vert\vert^{-2} \geq 0

    Does this seem right? I'm not sure how to manipulate it into the triangle inequality past this point.
    OK, you've got an error where you multiplied out

    "that complex expression" dot "that complex expression" (edit: I mean, v.v).

    If we expand out something of the form (a-b).(a-b) we get a^2 -2a.b + b^2 (which is what you've done).

    Your a.b term is incorrect.

    2nd edit: It took me a fair bit of looking to find the mistake; you might do better to simply expand it again and see if you get a different answer rather than looking. And since I might be away from keyboard for a bit, I'll spell it out in a spoiler:

    Spoiler:
    Show
    In the "a.b" term, where you have (y.y) you should have (x.y)
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    FWIW, the other "common method" for proving C-S is to go:

    ||x + \lambda y||^2 \ge 0 \quad \forall \lambda \in \mathbb{R}.

    so

     ||x||^2 + \lambda^2 ||y||^2 + 2\lambda (x \cdot y) \ge 0 \quad \forall \lambda (*). This is a quadratic in \lambda, and if its >=0 for all lambda we must have "B^2 <= 4AC". Do a bit of algebra and you get the required result.

    Now, the "cute" bit: if you differentiate (*), you can see it's minimized when \lambda = -(x \cdot y) / ||y||^2, and then x + \lambda y
    becomes the rather horrible expression you're given in the question...
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    Could prove Cauchy-Schwarz the 'traditional way' and expand both sides. There is also a clever (and quick) method involving considering the discriminant of the quadratic in m given by \displaystyle\sum_{i=1}^n (mx_i-y_i)^2.
    EDIT: Oh I see DFranklin has beat me to the punch with the cute solution. It really is a nice solution though, since you could prove it this way in your head in a matter of seconds.
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    (Original post by DFranklin)
    OK, you've got an error where you multiplied out

    "that complex expression" dot "that complex expression" (edit: I mean, v.v).

    If we expand out something of the form (a-b).(a-b) we get a^2 -2a.b + b^2 (which is what you've done).

    Your a.b term is incorrect.

    2nd edit: It took me a fair bit of looking to find the mistake; you might do better to simply expand it again and see if you get a different answer rather than looking. And since I might be away from keyboard for a bit, I'll spell it out in a spoiler:

    Spoiler:
    Show














    In the "a.b" term, where you have (y.y) you should have (x.y)













    Ah of course, a small slip up from getting confused with all the vertical lines probably and not noticing that the middle term is a multiple of the 3rd term in the expansion.

    So I have {\bf{v \cdot v}} = {\bf{\vert x \cdot x \vert}}-2 {\bf{\vert x \cdot y \vert^2 \vert \vert y \vert \vert^{-2}}}+{\bf{\vert x \cdot y \vert^2 \vert \vert y \vert \vert^{-2}}} \geq 0

    which leads to {\bf{\vert \vert x \vert \vert^2}} \geq {\bf{\vert x \cdot y \vert^2 \vert \vert y \vert \vert^{-2}}}

    then {\bf{\vert \vert x \vert \vert^2}} \cdot {\bf{\vert \vert y \vert \vert^2}} \geq {\bf \vert x \cdot y \vert^2}

    and taking the square root of both sides gives the required result.

    To be fair, I prefer the discriminant approach :lol:

    I have proven the triangle inequality, and I think I have got Q4 (which is marked as the second Q3) but I'm not sure if there is a different approach because the one I took uses an inequality that I don't think has any link to the previous part, or at least I'm unsure how to show the link.

    So {\bf \vert \vert x \vert \vert }^2=\vert x_1 \vert^2 + \vert x_2 \vert^2 + ... +\vert x_n \vert^2 \leq (\vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert)^2

    Then square rooting both sides gives the result, but I haven't proven that the sum of squares is always less than or equal to the square of the sum - I just used intuition and googled the proof which can be done by induction - but how can I use anything from the Q's above (if anything) to prove this??

    Also for Q5 I'm not quite sure how to start it off using Q4.

    Thanks for the help
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    (Original post by RDKGames)
    Ah of course, a small slip up from getting confused with all the vertical lines probably and not noticing that the middle term is a multiple of the 3rd term in the expansion.

    So I have {\bf{v \cdot v}} = {\bf{\vert x \cdot x \vert}}-2 {\bf{\vert x \cdot y \vert^2 \vert \vert y \vert \vert^{-2}}}+{\bf{\vert x \cdot y \vert^2 \vert \vert y \vert \vert^{-2}}} \geq 0

    which leads to {\bf{\vert \vert x \vert \vert^2}} \geq {\bf{\vert x \cdot y \vert^2 \vert \vert y \vert \vert^{-2}}}

    then {\bf{\vert \vert x \vert \vert^2}} \cdot {\bf{\vert \vert y \vert \vert^2}} \geq {\bf \vert x \cdot y \vert^2}

    and taking the square root of both sides gives the required result.

    To be fair, I prefer the discriminant approach :lol:

    I have proven the triangle inequality, and I think I have got Q4 (which is marked as the second Q3) but I'm not sure if there is a different approach because the one I took uses an inequality that I don't think has any link to the previous part, or at least I'm unsure how to show the link.

    So {\bf \vert \vert x \vert \vert }^2=\vert x_1 \vert^2 + \vert x_2 \vert^2 + ... +\vert x_n \vert^2 \leq (\vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert)^2

    Then square rooting both sides gives the result, but I haven't proven that the sum of squares is always less than or equal to the square of the sum - I just used intuition and googled the proof which can be done by induction - but how can I use anything from the Q's above (if anything) to prove this??
    I don't know you need to formally use induction, the square of the sum is "obviously" equal to the sum of the squares plus the sum of all "cross terms" |x_i||x_j| and these terms are all non-negative, so ignoring them can't hurt you.

    However, I'm pretty sure you were supposed to do it by saying that

    {\bf x} = (x_1, 0, 0, ..., 0) + (0, x_2, 0, 0, ...) + ... + (0, ..., 0, x_n) and then repeatedly use the triangle inequality.

    Also for Q5 I'm not quite sure how to start it off using Q4.
    I don't think you want to use Q4, you want to use C-S.

    If you make the right choice for 'y' it comes out in 1 line. (You probably also want to argue that "WLOG, we can assume no x_i is negative).
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    (Original post by DFranklin)
    I don't know you need to formally use induction, the square of the sum is "obviously" equal to the sum of the squares plus the sum of all "cross terms" |x_i||x_j| and these terms are all non-negative, so ignoring them can't hurt you.

    However, I'm pretty sure you were supposed to do it by saying that

    {\bf x} = (x_1, 0, 0, ..., 0) + (0, x_2, 0, 0, ...) + ... + (0, ..., 0, x_n) and then repeatedly use the triangle inequality.

    I don't think you want to use Q4, you want to use C-S.

    If you make the right choice for 'y' it comes out in 1 line. (You probably also want to argue that "WLOG, we can assume no x_i is negative).
    I see, I think my approach would be allowed nonetheless. I'll look into it another time.

    Then attempting Q5 if I let {\bf \vert \vert y \vert \vert} = \sqrt{n} \Rightarrow {\bf y } = \underbrace{(1,1,...,1)}_{n \text{times}} \in \mathbb{R}^n \Rightarrow {\bf \vert x \cdot y \vert}^2 = \vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert

    So using C-S I get {\bf \vert \vert x \vert \vert \cdot \vert \vert y \vert \vert} = \sqrt{n}{\bf \vert \vert x \vert \vert} \geq {\bf \vert x \cdot y \vert} = \sqrt{\vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert}

    but I'm unsure where to go from here since \sqrt{\vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert} \not\geq \vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert
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    (Original post by RDKGames)
    I see, I think my approach would be allowed nonetheless. I'll look into it another time.

    Then attempting Q5 if I let {\bf \vert \vert y \vert \vert} = \sqrt{n} \Rightarrow {\bf y } = \underbrace{(1,1,...,1)}_{n \text{times}} \in \mathbb{R}^n \Rightarrow {\bf \vert x \cdot y \vert}^2 = \vert x_1 \vert + \vert x_2 \vert + ... +\vert x_n \vert
    The last bit is wrong, it's just x.y, not (x.y)^2 (and so your later concerns don't arise). (You also have an assumption that the x_i are all non-negative for this to work).
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    (Original post by DFranklin)
    The last bit is wrong, it's just x.y, not (x.y)^2 (and so your later concerns don't arise). (You also have an assumption that the x_i are all non-negative for this to work).
    Awesome, just been using the equation in Q1 for some reason and now I realised I needed to use the one in Q2 hence the square that tripped me up. Got the inequality to work now, thanks!

    Also, why must x_i \geq 0 WLOG for this to work if we are taking the modulus of these values anyhow?
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    (Original post by RDKGames)
    Also, why must x_i \geq 0 WLOG for this to work if we are taking the modulus of these values anyhow?
    Because x.y = \sum x_i y_i, NOT \sum |x_i y_i|. For your choice of y, this means x.y = \sum x_i, not \sum |x_i| as you want.

    e.g. (1, 1).(1, -1) = 0.

    To fix this, you can either argue that WLOG the x_i are >=0, or instead of picking y = (1, 1, ..., 1), you can pick each y_i to be 1 if x_i >=0 and -1 otherwise. Then |y| is still sqrt(n), but you now get x.y =\sum |x_i|
 
 
 
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