Show the Cauchy-Schwarz inequality holds on a Hilbert space How would one go about showing this? Its a question in one of the workbooks but it doesn't provide an answer. Any help would be appreciated.
 A: Somehow, on the whole internet, it seems that the simplest proof of Cauchy-
Schwarz has yet to be recorded.  At least I couldn't find it after several minutes of searching...  The most prominent is certainly the proof mentioned by Daniel Fischer in this comment above, but that always seemed quite contrived to me.  Here is the ``best'' proof imho:
let $x,y$ be unit vectors.
Then $\langle x-y,x-y \rangle = |x|^2-2\langle x,y\rangle+|y|^2 \geq 0$
so $\langle x,y \rangle \leq 1$
Now for any two nonzero vectors, $x,y$ (if one is $0$ the result is trivial), we have that
$\left\langle \frac{x}{|x|},\frac{y}{|y|}\right\rangle \leq 1$ by the result above.
So $\langle x,y \rangle \leq |x||y|$
Of course, we also need to show that $\langle x,y \rangle \geq -|x||y|$, but I will leave it to you to see how to modify the argument to obtain this inequality.
A: I am not sure what is  the "best" proof of this famous inequality, but I found the following remark helpful, at least conceptually.  
For a 2 dimensional Hilbert space, i.e. the usual Euclidean plane of highschool math, the inequality is quite elementary and intuitive, by some drawing, or even working in coordinates, it is straighfword to show that $(ac+bd)^2\leq (a^2+b^2)(c^2+d^2)$. 
Now the remark is that for a general pair of vectors in a whatever dimension Hilbert space, you can consider the two dimensional subspace spanned  by these two vectors (supposing they are not linearly dependent, in which case there is nothing to prove). We are reduced to the previous case. 
In other words, it is really a plane geometry inequality, nothing more. 
A: As the Gramian matrix of $(x,y)$, namely $G(x,y):=\begin{pmatrix}\langle x,x\rangle & \langle x,y\rangle\\ \langle y,x\rangle & \langle y,y\rangle\end{pmatrix}$, is well known to be positiv-semidefinit, we know that $\det\bigl(G(x,y)\bigr)\ge0$ and equality holds iff $x\parallel y$.
