Hölder inequality in case $q=p=2$. It should return the Cauchy-Schwarz inequality, but I'm having trouble with comparing the left sides of the inequalities: 
For example, if $x=(4,3)$ and $y=(3,-4)$, then 
$\sum_{v=1}^2 |x_vy_v| = 24$
(left side of Hölder) is not the same as 
$|\langle x_v,y_v \rangle|=0$
(left side of CS).
What am I missing?
 A: In your example, $$\langle x, y \rangle = \sum_{v=1}^n x_v y_v$$
So
$$|\langle x,y \rangle| = \left| \sum_{v=1}^n x_v y_v \right|$$
Therefore,
$$|\langle x,y \rangle| = \left| \sum_{v=1}^n x_v y_v \right| \leq \sum_{v=1}^n |x_v y_v| \leq \left(\sum_{v=1}^n |x_v|^p\right)^{1/p} \left(\sum_{v=1}^n |x_v|^q\right)^{1/q}$$
where the first inequality is the triangle inequality, and the second is Hölder's. Ignoring everything between the LHS and RHS, we get
$$|\langle x,y \rangle| \leq \left(\sum_{v=1}^n |x_v|^p\right)^{1/p} \left(\sum_{v=1}^n |x_v|^q\right)^{1/q}$$
Setting $p=q=2$ gives the special case of Cauchy-Schwarz.
A: Typically, we refer the Cauchy-Schwarz inequality to
$$ (\sum_{i=1}^n x_i \bar{y_i})^2 \leq (\sum_{i=1}^n |x_i|^2) (\sum_{i=1}^n |y_i|^2) $$
Note that there are no modulus inside the summation on the left hand sign.
If we regard $\mathbb{C}^n$ as an inner product space, the inequality can be written as
$$ (x,y)^2 \leq \|x\|^2 \|y\|^2 = (x,x) (y,y)$$
A: \begin{align}
\text{LHS of Schwarz}=|\langle x,y\rangle|=&\left|\sum x_vy_v\right|,\\
\text{LHS of Hölder}=\|xy\|_1=&\,\sum|x_vy_v|.
\end{align}
