# Cauchy-Schwarz inequality and min/max

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 $u$.
• Is the notion of u=v being the maximum a geometric one or an algebraic one? That is, why do we discover that u=v is the maximum (I don't mean just trying u=v after the C-S and see that it exists and is $\lvert\lvert u \rvert \rvert$)? It follows from the vector angle formula right? Sep 4 '15 at 14:32
• You can give both types of explanations. Geometric explanation that the inner product is like cosine of angle between two vectors which is maximized when they are parallel. And the it follows algebraically from the fact that this inequality is derived using $||a-\lambda b||^2>0$ and where the equality occurs when $a=\lambda b$. Sep 4 '15 at 23:38