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Thread: Spivak's Calculus - Schwarz Inequality

  1. #1
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    Post Spivak's Calculus - Schwarz Inequality

    19. (a) Prove that if $\displaystyle x_1 = ny_1$ and $\displaystyle x_2 = ny_2$ for some number $\displaystyle n$, then equality holds in the Schwarz inequality. Prove the same thing if $\displaystyle y_1 = y_2 = 0$. Now suppose that $\displaystyle y_1$ and $\displaystyle y_2$ are not both $\displaystyle 0$, and that there is no number $\displaystyle n$ such that $\displaystyle x_1 = ny_1$ and $\displaystyle x_2 = ny_2$. Then

    $\displaystyle 0 < (ny_1 - x_1)^2 + (ny_2 - x_2)^2 $
    $\displaystyle = n^2(y_1^2 + y_2^2) - 2n(x_1y_1 + x_2y_2) + (x_1^2 + y_2^2).$

    Using Problem 18, complete the proof of the Schwarz inequality (i.e. prove that $\displaystyle x_1y_1 + x_2y_2 <= \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$).

    I've proven the first two assumptions so far (by substitution), but I'm not sure how to prove the third. I know that I have to complete the square (which I've done in problem 18), but I can't seem to do that in this problem since there are double the variables. I've tried manipulating the inequality such that $\displaystyle x_1y_1 + x_2y_2 < \frac{1}{2n}(x_1^2 + y_2^2) + \frac{n}{2}(y_1^2 + y_2^2)$, but that doesn't seem to work either.
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Spivak's Calculus - Schwarz Inequality

    Quote Originally Posted by RogueDemon View Post
    19. (a) Prove that if $\displaystyle x_1 = ny_1$ and $\displaystyle x_2 = ny_2$ for some number $\displaystyle n$, then equality holds in the Schwarz inequality. Prove the same thing if $\displaystyle y_1 = y_2 = 0$. Now suppose that $\displaystyle y_1$ and $\displaystyle y_2$ are not both $\displaystyle 0$, and that there is no number $\displaystyle n$ such that $\displaystyle x_1 = ny_1$ and $\displaystyle x_2 = ny_2$. Then

    $\displaystyle 0 < (ny_1 - x_1)^2 + (ny_2 - x_2)^2 $
    $\displaystyle = n^2(y_1^2 + y_2^2) - 2n(x_1y_1 + x_2y_2) + (x_1^2 + y_2^2).$

    Using Problem 18, complete the proof of the Schwarz inequality (i.e. prove that $\displaystyle x_1y_1 + x_2y_2 <= \sqrt{x_1^2 + x_2^2}\sqrt{y_1^2 + y_2^2}$).

    I've proven the first two assumptions so far (by substitution), but I'm not sure how to prove the third. I know that I have to complete the square (which I've done in problem 18), but I can't seem to do that in this problem since there are double the variables. I've tried manipulating the inequality such that $\displaystyle x_1y_1 + x_2y_2 < \frac{1}{2n}(x_1^2 + y_2^2) + \frac{n}{2}(y_1^2 + y_2^2)$, but that doesn't seem to work either.


    If I followed your question properly, then in my opinion discriminant of quadratic equation would help you.
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    Re: Spivak's Calculus - Schwarz Inequality

    Do you know how to apply the discriminant to completing this proof? I can't figure it out. :/
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Spivak's Calculus - Schwarz Inequality

    Quote Originally Posted by RogueDemon View Post
    Do you know how to apply the discriminant to completing this proof? I can't figure it out. :/
    $\displaystyle n^2(y_1^2 + y_2^2) - 2n(x_1y_1 + x_2y_2) + (x_1^2 + y_2^2)>0$

    $\displaystyle n^2a - 2nb + c>0$ - Polynomial of degree 2.

    $\displaystyle \Delta =b^2-4ac<0$(Why?)
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    Re: Spivak's Calculus - Schwarz Inequality

    Quote Originally Posted by Also sprach Zarathustra View Post
    $\displaystyle n^2(y_1^2 + y_2^2) - 2n(x_1y_1 + x_2y_2) + (x_1^2 + y_2^2)>0$

    $\displaystyle n^2a - 2nb + c>0$ - Polynomial of degree 2.

    $\displaystyle \Delta =b^2-4ac<0$(Why?)
    By completing the square, it becomes evident that $\displaystyle n^2a - 2nb + c = n + \frac{-b + \sqrt{-(b^2 - ac)}}{a}$. Thus, it must be true that $\displaystyle b^2 - 4ac < 0$ since the contrary would imply that the expression is complex.

    Though, how does this complete the proof of the Schwarz inequality?
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    Re: Spivak's Calculus - Schwarz Inequality

    Wait, nvm, I got it. Thanks!
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