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The proof given for the inequality in my lecture notes seems confusing to me. The proof is as follows:

Let $\lambda$ be any number and consider the vector $\lambda\textbf{v}-\textbf{w}$. Since $||\lambda\textbf{v}-\textbf{w}||\geqslant 0$ we have $$||\lambda\textbf{v}-\textbf{w}||^2\geqslant 0$$ $$\therefore (\lambda\textbf{v}-\textbf{w},\lambda\textbf{v}-\textbf{w})\geqslant 0$$ $$\therefore {\lambda}^2(\textbf{v},\textbf{v})-2\lambda(\textbf{v},\textbf{w})+(\textbf{w},\textbf{w})\geqslant 0$$ $${\lambda}^2-\frac{2\lambda(\textbf{v},\textbf{w})}{||\textbf{v}||^2}+\frac{||\textbf{w}||^2}{||\textbf{v}||^2}\geqslant 0$$ Completing the square gives: $$\left(\lambda-\frac{(\textbf{v},\textbf{w})}{||\textbf{v}||^2}\right)^2-\left(\frac{(\textbf{v},\textbf{w})}{||\textbf{v}||^2}\right)^2+\frac{||\textbf{w}||^2}{||\textbf{v}||^2}\geqslant 0$$

Now the proof sets $$\lambda=\frac{(\textbf{v},\textbf{w})}{||\textbf{v}||^2}$$ And proceeds to get the inequality. I understand that since $\textbf{v}$ and $\textbf{w}$ can be any vectors this gives the fact that $\lambda$ can be any number. However why does $\lambda$ always $\textbf{have}$ to be equal to the fraction above.

i.e. Surely I can pick a $\textbf{v},\textbf{w}$ and $\lambda$ such that $$\lambda-\frac{(\textbf{v},\textbf{w})}{||\textbf{v}||^2}\neq 0$$

Any help would be appreciated. Thanks.

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  • $\begingroup$ This value is chosen simply because the proof follows immediately after. $\endgroup$ – Brad Mar 30 '14 at 18:31
  • $\begingroup$ Other proofs of Cauchy-Schwarz are discussed here. math.stackexchange.com/questions/436559/… $\endgroup$ – littleO Mar 30 '14 at 18:35
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The very fact mentioned in the beginning, that $\lambda$ can be any number, allows us to pick out $$\lambda=\dfrac{\langle\mathbf{v,w}\rangle}{||\mathbf{v}||^2}$$ Of course you can pick out $\mathbf{v,w},\lambda$ such that $$\lambda\neq\dfrac{\langle\mathbf{v,w}\rangle}{||\mathbf{v}||^2}$$ However, this will in turn result in some number $\epsilon$, which can be any number (as $\lambda$ is any number and picking out $\mathbf{v,w}$ gives you a specific value for $\dfrac{\langle\mathbf{v,w}\rangle}{||\mathbf{v}||^2}$), which in turn you can set to $0$, due to its arbitrariness. This allows you to conclude the result.

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  • $\begingroup$ AH ok so you will end up with $||\textbf{v}||^2||\textbf{w}||^2\geqslant (\textbf{v},\textbf{w}) -k, ;\ k\geqslant 0$. Thus the inequality will still stand as the right hand side is still smaller than the left. So we can just set $k=0$ for simplicity. $\endgroup$ – George1811 Mar 30 '14 at 18:45
  • $\begingroup$ @George1811 Yes, exactly. $\endgroup$ – user122283 Mar 30 '14 at 18:48
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The derivation is correct for any value of $\lambda$. Choosing one precise $\lambda$ let you to conclude the desired inequality.

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