The choice of scalar factors in the proof of the Schwarz inequality
In this proof for the Schwarz Inequality, they seemingly arbitrarily choose $r = w\cdot w$ and $s =-(v\cdot w)$. Why did they make these selections? I don't understand where these values for $r$ and $s$ come from. Please guide.
1 Answer
$\begingroup$This proof is unnecessarily complicated. It's easier to start with
$$ \|\mathbf v+s\mathbf w\|^2\ge0\;, $$
which yields
$$ \mathbf v\cdot\mathbf v+2s\mathbf v\cdot\mathbf w+s^2\mathbf w\cdot\mathbf w\ge0\;, $$
and then differentiate with respect to $s$ to find the "worst case" of $s$, yielding
$$ 2\mathbf v\cdot\mathbf w+2s\mathbf w\cdot\mathbf w=0\;. $$
Since the claim is trivially true for $\mathbf w=\mathbf 0$, we can solve for $s$,
$$ s=-\frac{\mathbf v\cdot\mathbf w}{\mathbf w\cdot\mathbf w}\;, $$
and subtitute above, which, after multiplying through by $\mathbf w\cdot\mathbf w$ and simplifying, yields the claim.
You can do something similar in the two-variable case – differentiating with respect to $s$ yields
$$ 2r\mathbf v\cdot\mathbf w+2s\mathbf w\cdot\mathbf w=0\;, $$
and then the choice $r=\mathbf w\cdot\mathbf w$ and $s=-\mathbf v\cdot\mathbf w$ seems less magical.
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