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Homework Help: The Cauchy-Schwarz inequality for real numbers

  1. Jan 12, 2005 #1
    Ok...here is some back ground into my new found situation. I have done very well in every math class up to this point in time so I felt it was time for me to start looking at taking some more difficult classes. That being said I am technically still in my freshman year in college so I may have bitten off more than I can chew by taking a senior level class on inequalities. This is a special class in which we will be studying out the book called, "The Cauchy-Schwarz Master Class." There are all of 8 people in class who are all seniors and have all at least taken linear algebra except of course me, where the two highest math classes I have taken so far are Calculus 2 and Elementary set and logic Theory. Today was our first day of class and after discussing the syllabus and other general formalities we proceeded with a proof. I did my best to follow everything in class but I felt a little bit lost and so I am going to attempt to work through the proof here and I will ask questions as we go.

    The inequality in question is

    a_1b_1+a_2b_2+\cdots+a_nb_n \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}

    While it is easy to show that this is true for n=1

    \vert a_1b_1\vert\leq\sqrt{a_1^2b_1^2}

    It is a bit harder with n=2.

    Here we proceed in an algebraic manner until we get somewhere we know it is true.

    \vert a_1b_1+a_2b_2\vert\leq\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}

    Squaring both sides of the inequality is legal so long as both sides of the inequality are positive. So after squaring we have

    (a_1b_1)^2+2a_1b_1a_2b_2+(a_2b_2)^2 \leq\ (a_1b_1)^2+(a_1b_2)^2+(a_2b_2)^2+(a_2b_2)^2

    So far I am good with everything that is being done. The inequality is then simplified and we end up with

    2a_1b_1a_2b_2\leq\ (a_1b_2)^2+(a_2b_1)^2

    moving the LHS to the RHS and then reversing the entire inequality we ended up with this

    (a_1b_2)^2-2a_1b_1a_2b_2+(a_2b_1)^2\geq\ 0

    Which factors into a perfect square which is always either positive or equal to zero and so this last inequality is always true and so we now have that Cauchy's inequality is true of n=2.

    This is the basis step in our induction proof.
    Our induction hypothesis is that the Cauchy inequality is true for some n so we have

    a_1b_1+a_2b_2+\cdots+a_nb_n \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}

    we must show that it is true of n+1, namely we must show that
    a_1b_1+a_2b_2+\cdots+a_nb_n+a_{n+1}b_{n+1} \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2+a_{n+1}^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2+b_{n+1}^2}

    Before we go any farther I would like it if perhaps some one could help me out a bit on induction proofs. I have no problem understanding why an induction proof works when we show it true for n=1 and then show that if it is true for some n it is also true for n+1 and then conclude that it is therefore true for all natural numbers. Where I don't really follow the logic is when our basis step is a number other than 1, like in this case where n=2.
    Secondly, is it true that induction works up and down? In other words when my basis step shows something true for n=1 and I show that if it is true for n then it is true for n+1, does this mean that it is true for all integers? Or only integers greater than 1.

    To keep the over all post size down I will continue this on another post
    To be continued...
    Last edited: Jan 12, 2005
  2. jcsd
  3. Jan 12, 2005 #2

    Ok moving along (I wish making tex code was easier to do)
    To show that if this thing is true for n, then it is true for n+1 we need a good tool in our repertoire, namely the triangle inequality which says

    \vert a+b \vert \leq \vert\ a \vert + \vert b \vert

    Now we do some algebra and try to proceed where ever we might find a way to do so. Knowing that we need to use our induction hyposthesis we need to start by finding a way to break up the expression

    \vert a_1b_1+a_2b_2+\cdots+a_nb_n+a_{n+1}b_{n+1} \vert

    Some grouping and the triangle inequality and our induction hypothesis allows us to say

    \vert (a_1b_1+a_2b_2+\cdots+a_nb_n)+(a_{n+1}b_{n+1}) \vert \leq \vert a_1b_1+ \cdots +a_nb_n \vert + \vert a_{n+1}b_{n+1} \vert \leq \sqrt{a_1^2+a_2^2+\cdots+a_n^2} \sqrt{b_1^2+b_2^2+\cdots+b_n^2}+ \vert a_{n+1}b_{n+1} \vert

    Ok, now I am confused....

    I am suppose to use our previous result that the inequality is true for n=2 to get a new inequality but no matter how many times I read my notes or the book I am not seeing what is happening. I feel like there is just some magic hand waving going on here but I know there is a reason and it just evades me. If anyone can see where I need to go from here please give me hint.

    Last edited: Jan 12, 2005
  4. Jan 12, 2005 #3


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    To answer, I will quote you.

    The basis step was not to show that it is true for n=2; it was to show it is true for n=1. Showing it is true for n=2 has nothing to do with the logic behind the principle of the proof by induction, it has been done only so you can use this result in your proof. Kind of like a Lemma, if you like.

    Yes, only intergers greater than 1. Because the theorem you are trying to prove state that it is true for all positive integers. Therefor, when you say "suppose it is true for n", you mean "suppose it is true for some positive integer n", not "suppose it is true some some integer n".

    However, I don't see why it wouldn't be true that you have proven the theorem for all integer lesser than z if you show it is true for z and suppose it is true for and integer b lesser than z and then proceed to show it implies that it is true for b-1.

    As an exercise to make sure you understand the idea of induction correctly: "Suppose we have shown that for proposition P, P is true for some integer n. We then proceed to show that if it is true for m lesser or equal to n, it is also true for m+1. Given z lesser than n, determine the numbers a,b delimiting the range of integers [itex][a,b]\cap \mathbb{Z}[/itex] for which P is true, according to the principle of induction."
  5. Jan 12, 2005 #4


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    As for your main question, I don't see how to continue either. But perhaps you would increase your chance of getting help on your problem by posting what is written in your book or in your notes after that last inequality. That way we wouldn't have to find a way to continue but just to understand the proof.
    Last edited: Jan 12, 2005
  6. Jan 12, 2005 #5
    Good point..

    the next inequality is

    \sqrt{a_1^2+a_2^2+ \cdots +a_n^2} \sqrt{b_1^2+b_2^2+ \cdots +b_n^2} + \vert a_{n+1}+b_{n+1} \vert \leq \sqrt{a_1^2+a_2^2+ \cdots +a_n^2+a_{n+1}^2} \sqrt{b_1^2+b_2^2+ \cdots +b_n^2+b_{n+1}^2}

    And thus the proof is done because this is what is desired. I just cannot see how to come to this conclusion...

    The only explanation for this is that is used the previous result that the inequality is true for n=2. I think I will sleep on this one for tonight as I am kind of getting burned out for today. Thanks for you help so far quasar987.

  7. Jan 13, 2005 #6


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    I believe you mean :

    \sqrt{a_1^2+a_2^2+ \cdots +a_n^2} \sqrt{b_1^2+b_2^2+ \cdots +b_n^2} + \vert a_{n+1}b_{n+1} \vert \leq \sqrt{a_1^2+a_2^2+ \cdots +a_n^2+a_{n+1}^2} \sqrt{b_1^2+b_2^2+ \cdots +b_n^2+b_{n+1}^2}

    It's easy to show how this is true if you make the following substitutions to help visualize :

    [tex] M = {a_1^2+a_2^2+ \cdots +a_n^2} [/tex]
    [tex]N= {b_1^2+b_2^2+ \cdots +b_n^2} [/tex]
    [tex] m = a_{n+1}[/tex]
    [tex] n = b_{n+1} [/tex]

    Then this last step reduces to the statement :

    [tex]\sqrt {MN} + |mn| \leq \sqrt {(M+m^2)}\sqrt{(N+n^2)} [/tex]

    Now does it look familiar ?
    Last edited: Jan 13, 2005
  8. Jan 13, 2005 #7

    Here's a version of the proof found in older algebra textbooks:

    The Cauchy Schwarz inequality states that

    [tex]a_{1}b_{1} + a_{2}b_{2} + ..... + a_{n}b_{n} \leq \sqrt{a_{1}^2 + a_{2}^2 + .... + a_{n}^2}\sqrt{b_{1}^2 + b_{2}^2 + .... + b_{n}^2}[/tex]

    Proving it is equivalent to proving the inequality [tex]C \leq \sqrt{A}\sqrt{B}[/tex]

    [tex]A = a_{1}^2 + a_{2}^2 + .... + a_{n}^2 = \sum_{i = 1}^{n}a_{i}^2[/tex]
    [tex]B = b_{1}^2 + b_{2}^2 + .... + b_{n}^2 = \sum_{i = 1}^{n}b_{i}^2[/tex]
    [tex]C = a_{1}b_{1} + a_{2}b_{2} + ..... + a_{n}b_{n} = \sum_{i = 1}^{n}a_{i}b_{i}[/tex]

    Next consider the realvalued function f(t) defined by,

    [tex]f(t) = \sum_{i = 1}^{n}(a_{i}t + b_{i})^{2}[/tex]

    Expanding it gives you

    [tex]f(t) = At^{2} + 2Ct + B[/tex]

    By the very construction, f(t) >= 0 for all real t. Also, A > 0 (if A = 0 then the inequality holds trivially) so the only way both these constraints can be satisfied is that the discriminant of the equation f(t) = 0 be nonpositive, i.e.

    [tex]4C^2 - 4AB \leq 0[/tex]

    which is what we set out to prove!

    I am sorry if this isn't germane to the discussion...I just thought I'd pop it in (though I know you may already have done this). It might be useful to those new to the Cauchy Schwarz "phenomenon" :-).

  9. Jan 13, 2005 #8
    I got it now, whew, this just uses our previously established inequality for n=2, just like the book said. This really makes me mad when I stare at a problem for so long and the answer is literally staring right back at me but I cannot see it. Well I appreciate the help everyone.

    As I progress through this course I plan to use this thread to ask any questions I might be stuck on.

  10. Jan 18, 2005 #9
    Edit: Oops... didn't realize this was for real numbers. Was looking at it in the context of complex numbers.
    Last edited: Jan 18, 2005
  11. Jan 23, 2005 #10
    I have a couple more questions I need to ask about this inequality. To being with, I want to be able to make sure everyone understands exactly what I am talking about and so to avoid me screwing up any kind of translation I will try to copy the text verbatim. Then I can explain what parts do not sit well with me and then hopefully someone would be so kind and help me out a bit.

    Starting with:

    One of the most immediate qualitative inferences from Cauchy's inequality is
    the simple fact that

    [tex]\sum_{k=1}^{\infty}{a_k^2} < \infty \mbox{ And } \sum_{k=1}^{\infty}{b_k^2} < \infty \mbox{ implies that } \sum_{k=1}^{\infty}{|a_kb_k|} < \infty[/tex]

    Give a proof of this assertion that does not call on Cauchy's inequality.

    When we consider this challenge, we are quickly drawn to the realization that we need to show that the product [tex]a_kb_k[/tex] is small when [tex]a_k^2[/tex] and [tex]b_k^2[/tex] are small. We could be sure of this inference if we could prove the existence of a constant C such that

    [tex] xy \leq C(x^2+y^2)[/tex]


    Ok, this does not sit well with me. I am not an expert on infinite series and so there are a couple of jumps that seem too vague for me to just accept as true. The first thing I need to understand is what does it mean to say that an infinite series is less than infinity. When I try to answer this question this is what I come up.

    I know that the sequence [tex] \{a_1,a_2, \cdots ,a_n \} [/tex] may or may not converge as n goes to infinity but either way the series could still be less than infinity. So we are assuming that the series is less than infinity, but does that mean that it necessarily converges? I mean, couldn't the series have an upper bound and a lower bound? There is no assumption that our sequence of partial sums is monotonic so this thing can move up and down all it wants to and still be less than infinity, right?

    So all I can say is that the sequence of partial sums will always be less than or equal to the absolute value of some number, say x. so we have

    [tex]\sum_{k=1}^{\infty}{a_k^2} \leq |x| \mbox{ and of course } \sum_{k=1}^{\infty}{b_k^2} \leq |y|[/tex]

    Now exactly how does this imply that

    [tex]\sum_{k=1}^{\infty}{|a_kb_k|} < \infty[/tex]

    Well according to the text book, the excerpt from the text is given above, we need to show that the product [tex]a_kb_k[/tex] is small when [tex]a_k^2[/tex] and [tex]b_k^2[/tex] are small. It then says we could be sure of this inference if...

    Ok I would be cool with that if I knew for sure that the partial sums were converging on some number. In other words do we know that if

    [tex] S_2=a_1^2+a_2^2 [/tex]
    [tex] \vdots [/tex]
    [tex] S_n=a_1^2+a_2^2+\cdots +a_n^2} [/tex]
    [tex] \mbox{ then } S_n < S_{n+1}\ \forall n \in \mathbb{R}[/tex]

    And also that

    [tex]S_n \leq P \mbox{ for some real number P } \forall n \in \mathbb{R}[/tex]

    I am not even sure if that makes sense. What I am trying to ask in other words is, is the squence of partial sums a bounded monotonic sequence?
    If I knew that was true then I think it would be approiate to follow that line of reasoning. But do we know that?

    Thanks to anyone who will offer some advice.

  12. Jan 23, 2005 #11
    Can anyone offer me any advice on this problem?

    Is there something that is unclear? I don't know but I am certainly at a stand still and there are not a lot of places to get help on this kind of problem.


  13. Jan 23, 2005 #12
    Ok, I somehow do not see me getting any response with the way the question has been asked. So let me ask it in a much easier way that hopefully someone here can answer for me.

    What I need to know is what exactly does it mean to say that an infinite series is less than infinity? Does it mean that the limit of partial sums converges? Or can the partial sums never converge but also never blow up?

    Once more for good luck, can an infinite series never converge and always be less than some upper bounds?

    Thank you so much for reading my post and please if you can offer any suggestions I would really appreciate them.

    Best Regards
  14. Jan 23, 2005 #13


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    This seems unnecessarily complicated. Don't we know that

    [tex]|a_kb_k|\leq a_k^2+b_k^2[/tex] for any two integers [tex]a_k[/tex] and [tex]b_k[/tex]

    So doesn't it automatically follow that:

    [tex]\sum_{k=1}^{\infty}{|a_kb_k|}\leq \sum_{k=1}^{\infty}{a_k^2} + \sum_{k=1}^{\infty}{b_k^2}[/tex]

    So since the right hand side is <= infinity, so is the left hand side. Perhaps I missed something.
  15. Jan 23, 2005 #14
    I don't know...do we know that to be true? Let me work on that one for a while and see where it takes me.

    Thanks a bunch...
  16. Jan 24, 2005 #15


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    Yes, I'm pretty sure we know that.

    There are 3 possibilites: |ak|=|bk| or |ak|>|bk| or |ak|<|bk|

    If [tex]|a_k|=|b_k|[/tex]then (EDIT here included absolute value signs)

    [tex]|a_kb_k|=a_k^2 \leq a_k^2+b_k^2[/tex]

    I'll assume |bk|<|ak| now with no loss of generality

    [tex]|b_k|<|a_k| [/tex]
    [tex]|a_kb_k| < a_k^2[/tex]
    [tex]|a_kb_k|< a_k^2 + b_k^2[/tex]
    Last edited: Jan 24, 2005
  17. Jan 24, 2005 #16
    Yes, you are right...
    very direct and also very clear....I like it a lot.
  18. Jan 24, 2005 #17


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    Thanks. :smile:
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