# Spivak problem on Schwarz inequality

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 $\lambda$?

• 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 $|x_1 y_2 + x_2 y_2|$. – Calvin Lin May 15 '13 at 3:34

That is a typo. You need $\lambda\ge 0$.
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 $x_1y_1 + x_2y_2\leq \left|x_1y_1 + x_2y_2\right|$.
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 $\lambda > 0$