Spivak's Calculus - Schwarz Inequality

**RogueDemon** · July 14th 2011, 01:10 PM

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

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

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

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 $x_1y_1 + x_2y_2 < \frac{1}{2n}(x_1^2 + y_2^2) + \frac{n}{2}(y_1^2 + y_2^2)$ , but that doesn't seem to work either.

**Also sprach Zarathustra** · July 14th 2011, 02:51 PM

Originally Posted by RogueDemon

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

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

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

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 $x_1y_1 + x_2y_2 < \frac{1}{2n}(x_1^2 + y_2^2) + \frac{n}{2}(y_1^2 + y_2^2)$ , 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.

**RogueDemon** · July 14th 2011, 04:35 PM

Do you know how to apply the discriminant to completing this proof? I can't figure it out. :/

**Also sprach Zarathustra** · July 14th 2011, 04:42 PM

Originally Posted by RogueDemon

Do you know how to apply the discriminant to completing this proof? I can't figure it out. :/

$n^2(y_1^2 + y_2^2) - 2n(x_1y_1 + x_2y_2) + (x_1^2 + y_2^2)>0$

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

$\Delta =b^2-4ac<0$ (Why?)

**RogueDemon** · July 14th 2011, 05:27 PM

Originally Posted by Also sprach Zarathustra

$n^2(y_1^2 + y_2^2) - 2n(x_1y_1 + x_2y_2) + (x_1^2 + y_2^2)>0$

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

$\Delta =b^2-4ac<0$ (Why?)

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

Though, how does this complete the proof of the Schwarz inequality?

**RogueDemon** · July 14th 2011, 05:31 PM

Wait, nvm, I got it. Thanks!

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