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I have a question regarding problem 19 in the 3rd Ed. of Spivak's Calculus. Specifically, part (a). The question concerns the Schwarz inequality: $$ x_1y_1 + x_2y_2 \leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2} \ . $$ It says to prove that if $x_1=\lambda y_1$ and $x_2 = \lambda y_2$ for some number $\lambda$, then equality holds in the Schwarz inequality.

Substituting the given values for $x_1$ and $x_2$ we have $$ \lambda (y_1^2+ y_2^2) \leq |\lambda|(y_1^2+y_2^2) \ . $$ It appears to me that equality can only hold if $\lambda \geq 0$. Can someone explain to me how equality holds for any given $\lambda$?

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    $\begingroup$ It depends on which version you use. Some assume that all the variables are non-negative. Other versions use the absolute value. Considering that you have $|\lambda|$, you should also have $|x_1 y_2 + x_2 y_2|$. $\endgroup$ – Calvin Lin May 15 '13 at 3:34
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That is a typo. You need $\lambda\ge 0$.

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    $\begingroup$ Was this supposed to be an answer? $\endgroup$ – Inceptio May 15 '13 at 3:34
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    $\begingroup$ This fits much better as a comment indeed. $\endgroup$ – user1620696 May 15 '13 at 3:36
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    $\begingroup$ It answered the question. You guys have way too much free time on your hands. $\endgroup$ – Ted Shifrin May 15 '13 at 3:39
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    $\begingroup$ Seems to me like you do too, Ted, if you were compelled to spend time and give a non-answer ;-) $\endgroup$ – Namaste May 15 '13 at 3:48
  • $\begingroup$ @TedShifrin For what it is worth, that was a sufficient answer to me as I suspected a typo. $\endgroup$ – user59083 May 15 '13 at 4:18
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From what I recall, the actual statement of the inequality is $$ \left|\langle x,y \rangle\right|\leq \lVert x\rVert\cdot\lVert y\rVert $$ or, with your notations $$ \left|x_1y_1 + x_2y_2\right|\leq \sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2} $$ so the book is right (you'll have $|\lambda|$ on both sides). Note that this is a stronger statement than what you made, since we always have $x_1y_1 + x_2y_2\leq \left|x_1y_1 + x_2y_2\right|$.

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If you Chose to define $y_1 \geq x_1 $ and $ y_2 \geq x_2 $ without loss of generality then $ |\lambda | = \lambda $ since the choice is arbitrary in the formula you don't need to define $ \lambda > 0 $

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