Why use absolute value for Cauchy Schwarz Inequality? I see the Cauchy-Schwarz Inequality written as follows 
$$|\langle u,v\rangle| \leq \lVert u\rVert \cdot\lVert v\rVert.$$
Why the is the absolute value of $\langle u,v\rangle$ specified? Surely it is apparent if the right hand side is greater than or equal to, for example, $5$, then it will be greater than or equal to $-5$?
 A: I assumed that we work in a real inner product space, otherwise of course we have to put the modulus. 
The inequality $\langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$ is also true, but doesn't give any information if $\langle u,v\rangle\leq 0$, since in this case it's true, and just the trivial fact that a non-negative number is greater than a non-positive one. What is not trivial is that $\lVert u\rVert\lVert v\rVert$ is greater than the absolute value. But in fact the assertions 
$$\forall u,v \quad \langle u,v\rangle\leq \lVert u\rVert\lVert v\rVert$$
and 
$$\forall u,v\quad |\langle u,v\rangle|\leq \lVert u\rVert\lVert v\rVert$$
are equivalent. Indeed, the second implies the first, and consider successively $u$ and $-u$ in the first to get the second one.
A: What if the inner product is a complex number which can happen if $u$ and $v$ are vectors of
complex numbers?
For real vectors, the Cauchy-Schwarz Inequality is better written as
$$-||u||\cdot ||v|| \leq \langle u, v \rangle \leq ||u||\cdot ||v||$$
where, if $||v|| > 0$, then equality holds in the right inequality
if $u = \lambda v$ with $\lambda \geq 0$
and in the left inequality if  $u = \lambda v$ with $\lambda \leq 0$.
