The Cauchy-Schwarz inequality for vectors is:
$$|\langle u,v\rangle| \leq \lVert u\rVert \cdot\lVert v\rVert,$$
which holds even without the absolute value.
However when applying this to find e.g. the maximum of a function $f\colon\mathbb{R^2}\to\mathbb{R}$, $f(x,y)=\frac{3x-2y}{\sqrt{x^2+y^2}}$, where one sets $u=(3,-2)$ and $v=(x,y)$, I have $$ f(x,y)=\frac{3x-2y}{\sqrt{x^2+y^2}}=\frac{\langle u,v\rangle}{\lVert v\rVert} \leq\frac{\lVert u\rVert \cdot\lVert v\rVert}{\lVert v\rVert} = \lVert u\rVert $$
That is, using the C-S leads to $f(x,y)\leq\lVert u\rVert$, but is $\lVert u\rVert$ really the maximum value of this function?
Or more generally, does the C-S always display the maximum value, given any similar function?
yes. and the maximum is attained when coordinates of $x$ are proportional to the coordinates of $u$.