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I'm studying multivariable calculus. Usually, when I study, I go through a book until I find a theorem, and then try to prove it. I was unable to, so I read the proof, which is the following:

Let $x, y \in \mathbb{R}^m, \alpha \in \mathbb{R}$. Then $(x+\alpha y)\bullet(x+\alpha y) = \vert \vert x+\alpha y\vert\vert^2 \geq0$. Using the properties a the inner product we get:

$(x+\alpha y)\bullet(x+\alpha y) = x\bullet x+\alpha x\bullet y + \alpha y\bullet x + \alpha^2y\bullet y = \vert\vert x\vert\vert^2+2(x\bullet y)\alpha + \alpha²\vert\vert y\vert\vert^2 \geq 0$.

That last inequality is true iff the discriminant of the polynomial with respect to $\alpha$ is less than or equal to 0. Therefore $\vert x\bullet y\vert - \vert \vert x\vert\vert²\vert\vert y\vert\vert^2 \leq 0$, from which comes the Cauchy-Schwarz inequality. Q.E.D

I can follow every step of the proof. I also get the intuition of why the inequality should be true. However, the proof seems "empty" to me. I don't understand what someone who wanted to prove this would do to find it. What's the intuition behind using $x+\alpha y$?

The reason I ask this is because, after I read the proof, the way used to prove it was so beyond everything that I tried, that I am almost sure that I'd never be able to prove this on my own. How to deal with these kind of situations?

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    $\begingroup$ Cauchy-schwarz boils down to the fact that, if you project one vector onto the other, the square of the length of the perpendicular component is greater than or equal to $0$, see my answer here. $\endgroup$ – Theo Bendit Mar 23 at 14:12
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Proving theorems about general vector spaces, or general inner product spaces, can begin by considering a familiar $2$- or $3$-dimensional space. But then you need to abstract the intuition so it's pure algebra, now diagrams required. So your question comes down to what sort of preamble may have helped here.

If you think about vectors in a space you can visualise, all the theorem says is that the angle $\theta$ between two vectors satisfies $-1\le\cos\theta\le 1$, which by the cosine rule is equivalent to the triangle inequality. Since the cosine rule can be stated in terms of dot products, it makes sense to see what you learn from one more equivalent result, $\Vert x-y\Vert^2\ge 0$.

But $\Vert x-\alpha y\Vert^2\ge 0$ is a natural generalisation, and connects the issue to extremising quadratics, with the extremum giving us the most inequality we can get. And we don't need to think about a specific vector space to use $\Vert v\Vert^2\ge 0$, so it's a general starting point.

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  • $\begingroup$ Wouldn't any real angle $\theta$ satisfy $-1 \leq cos \theta \leq 1$ ? I'm not very familiar with cosine rule or geometric proofs though $\endgroup$ – RUBEN GONÇALO MOROUÇO Mar 23 at 11:58
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    $\begingroup$ @RUBENGONÇALOMOROUÇO It would, yes. One could argue that knowing that is equivalent to knowing every other theorem mentioned herein. $\endgroup$ – J.G. Mar 23 at 12:01
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I don't know about anybody else, but I share your dissatisfaction with the standard slick proof, and I personally find it helpful to think instead of expressing $x$ as a sum of a multiple of $y$ and a vector orthogonal to $y$. This kind of resolution of a vector into two mutually orthogonal components is a common and natural operation.

If $\lambda$ is real, then $x - \lambda y$ is orthogonal to $y$ if and only if (in your notation) $(x - \lambda y) \bullet y = 0$, i.e., $$ \lambda \|y\|^2 = x \bullet y. $$

For any value of $\lambda$ satisfying that condition ($\lambda$ may be chosen arbitrarily if $y = 0$, and there is a unique solution for $\lambda$ if $y \ne 0$), write $u = x - \lambda y$ and $v = \lambda y$, so that $x = u + v$ and $u \bullet v = 0$. Then: \begin{align*} \|x\|^2 & = (u + v) \bullet (u + v) \\ & = u \bullet u + 2u \bullet v + v \bullet v \\ & = \|u\|^2 + \|v\|^2 \\ & \geqslant \|v\|^2. \end{align*} Therefore, using the definitions of $v$ and $\lambda$: $$ \|x\|^2\|y\|^2 \geqslant \|v\|^2\|y\|^2 = \lambda^2\|y\|^4 = (x \bullet y)^2 = |x \bullet y|^2, $$ and the result follows. So the selection of the value $-\lambda$ for $\alpha$ does make some intuitive sense (to me, at least).

You could arrive at this value of $\alpha$ less intuitively by "completing the square" in the expression you derived for $\|x + \alpha y\|^2$, thus, multiplying by $\|y\|^2$, to avoid a possible division by zero: \begin{align*} \|x + \alpha y\|^2\|y\|^2 & = \|x\|^2\|y\|^2 + 2(x \bullet y)\alpha\|y\|^2 + \alpha^2\|y\|^4 \\ & = (\alpha\|y\|^2 + x \bullet y)^2 + \|x\|^2\|y\|^2 - (x \bullet y)^2 \\ & = \|x\|^2\|y\|^2 - (x \bullet y)^2, \end{align*} if $$\alpha\|y\|^2 + x \bullet y = 0. $$ So the proof you quoted can be seen as the proof by resolution into orthogonal components in heavy disguise.

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    $\begingroup$ (+1) for this approach $\endgroup$ – Mark Viola Mar 23 at 14:25
  • $\begingroup$ One can argue more directly that $\|x-\lambda y\|^2=(x-\lambda y)\bullet x=\|x\|^2-\lambda(x\bullet y)$, therefore $\|x\|^2\geqslant\lambda(x\bullet y)$, therefore $\|x\|^2\|y\|^2\geqslant\lambda\|y\|^2(x\bullet y)=(x\bullet y)^2$; but this loses the nice intuition of Pythagoras's theorem, and its direct corollary that the projection of $x$ on $y$ is shorter than $x$ (although this can be recovered by writing $\lambda(x\bullet y)$ as $\lambda^2\|b\|^2=\|\lambda b\|^2$); and the proof veers towards being, once again, "slick" and unmemorable - which is why I forgot having once done it this way! $\endgroup$ – Calum Gilhooley Mar 23 at 16:44

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