- #1

swevener

- 21

- 0

**The problem**

Given the Schwarz inequality, [itex]x_{1}y_{1} + x_{2}y_{2} \leq \sqrt{x_{1}^{2} + x_{2}^{2}} \sqrt{y_{1}^{2} + y_{2}^{2}}[/itex], prove that if [itex]x_{1} = \lambda y_{1}[/itex] and [itex]x_{2} = \lambda y_{2}[/itex] for some number [itex]\lambda \geq 0[/itex], then equality holds. Prove the same thing if [itex]y_{1} = y_{2} = 0[/itex]. Now suppose that [itex]y_{1}[/itex] and [itex]y_{2}[/itex] are not both 0 and that there is no number [itex]\lambda[/itex] such that [itex]x_{1} = \lambda y_{1}[/itex] and [itex]x_{2} = \lambda y_{2}[/itex]. Then

[tex]\begin{align*}

0 &\lt (\lambda y_{1} - x_{1})^{2} + (\lambda y_{2} - x_{2})^{2} \\

&= \lambda^{2} (y_{1}^{2} + y_{2}^{2}) - 2 \lambda (x_{1} y_{1} + x_{2} y_{2}) + (x_{1}^{2} + x_{2}^{2}).

\end{align*}[/tex]

Use the solutions to the quadratic equation to prove the Schwarz ineq.

**My confusion**

I can do all the parts of this, but I'm not sure how they fit together. I can't figure out how we go from the Schwarz ineq. to the quadratic equation, so I don't know why the lack of a real solution proves the ineq. I've tried working it forward and backward and all I've got is wasted paper and a sore wrist.