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My professor mentioned that reason $t$ was chosen during the Cauchy-Schwarz inequality proof can also be seen when we minimize $\lVert x\rVert^2 +2t\langle x,y \rangle +t^2\lVert y\rVert^2$ over $t \in \mathbb{R}$. I'm not really sure what he is getting at.

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up vote 4 down vote accepted

We know that $$ || x + ty||^2 \geq 0 $$ for all $t \in \mathbb{R}$. Then we have $$ 0 \leq ||x+ty||^2 = ||x||^2 + 2t\langle x, y\rangle + t^2||y||^2. $$

Hence the inequality $$ ||x||^2 + 2t\langle x, y\rangle + t^2||y||^2 \geq 0 $$ holds for all $t \in \mathbb{R}$. Now what is the "best" $t$ that we could choose? The smaller the left hand side is, the tighter a bound we have proven. This gives us the idea to minimize the left side in $t$. One may differentiate to find $$ 2\langle x, y \rangle + 2t ||y||^2 = 0 \implies t = \frac{-\langle x ,y \rangle}{||y||^2} $$ is where the minimum occurs (check that it is indeed a minimum using any method you prefer). If $||y||=0$ then $y=0$ and the Cauchy-Shwartz inequality holds anyway, so we exclude this case.

Plugging in that value of $t$ and rearranging then gives $\langle x, y \rangle^2 \leq ||x||^2 ||y||^2$ and taking square roots gives the Cauchy-Shwartz inequality.

Note that a much simpler way to derive this is to note that $$ 0 \leq ||x+ty||^2 = ||x||^2 + 2t\langle x, y\rangle + t^2||y||^2 $$ says that a parabola has at most one real root, so its disciminant $b^2-4ac$ must be nonpositive, i.e. $$ (2\langle x, y\rangle)^2 - 4 (||x||^2) (||y||^2) \leq 0 $$ which givees the desired inequality immediately.

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