Given $|a|$ and $|b|$, what is the smallest value of $|a + b|$? If a and b are vectors such that $|a| = 7$ and $|b| = 11$, then find the smallest possible value of $|a+b|$.
I thought it was $\sqrt{170}$, but its not true. Help! :(
 A: The norm of the sum of two vectors $\vec a$ and $\vec b$ can be written
$$\begin{align}
||\vec a+\vec b||&=\sqrt{\langle \vec a+\vec b,\vec a+\vec b\rangle}\\\\
&=\sqrt{||\vec a||^2+||\vec b||^2 +2\text{Re}\left(\langle \vec a,\vec b\rangle\right)} \tag 1\\\\
\end{align}$$
From the Cauchy-Schwarz Inequality for normed vector spaces 
$$|\langle \vec a,\vec b\rangle|\le ||\vec a||\,||\vec b||\,\tag 2$$
Using $(2)$ in $(1)$ provides upper and lower bounds for the  norm of the sum as
$$|\,||\vec a||\,-\,||\vec b||\,|\le\,||\vec a+\vec b||\,\le\,||\vec a||\,+\,||\vec b|| \tag 3$$
For the problem herein, $||\vec a||=7$ and $||\vec b||=11$.  Using $(3)$ yields
$$||7-11||\le||\vec a+\vec b||\le7+11\implies 4\le ||\vec a+\vec b||\le 18$$
Heuristically, we can see that the maximum and minimum of the norm of the sum $\vec a+\vec b$ simply occur for those cases for which $\vec a$ and $\vec b$ are parallel and anti-parallel, respectively.
A: If $\vec a$ and $\vec b$ are collinear and opposed to each other , then 
$|\vec a + \vec b| = 3$.
By the law of cosines,
$$
   |\vec a + \vec b|^2 =
   |\vec a|^2 + |\vec b|^2 - 2|\vec a||\vec b| \cos(\pi - \theta) =
   |\vec a|^2 + |\vec b|^2 + 2|\vec a||\vec b| \cos(\theta)
$$
where $\theta$ is the angle between the two vectors. this quantity is smallest when  $\theta = \pi$, where we get
$$
   |\vec a + \vec b|^2 =
   |\vec a|^2 + |\vec b|^2 - 2|\vec a||\vec b| =
   \left ( |\vec a| - |\vec b| \right)^2
$$
So the minimum value is $\left | |\vec a| - |\vec b| \right|$
