Vector Norm addition If a and b are vectors such that ||a||=7 and ||b||=11, then find the smallest possible value of ||a+b||.
So far I know that for a= $\begin{pmatrix} x \\ y \end{pmatrix}$,     $x^2 +y^2 = 49$ and for b= $\begin{pmatrix} m \\ n \end{pmatrix}$,     $m^2+n^2 =121$. 
What do I do now?
Thanks
 A: Since $||b||=||(a+b)+(-a)||\le||a+b||+||a||$, $\;\;||a+b||\ge||b||-||a||=4$.
This is the smallest value, since $b=-\frac{11}{7}a\implies||a+b||=\frac{4}{7}||a||=4$.
A: This question can be answered immediately in one's head, just by visualizing.  Visually, to make $\| a + b\|$ as small as possible, $a$ and $b$ should point in opposite directions.  In this case, $\| a + b \| = 4$.
@user84413 's solution shows how to convert this visual intuition into a rigorous proof.
A: It appears you are dealing with inner product spaces here. $||a+b||^2 = ||a||^2 + ||b||^2 + 2 \langle a,b\rangle \geq ||a||^2 + ||b||^2 - 2||a||||b|| = 7^2 + 11^2 - 2\times7\times 11 = 4^2$. The inequality invoked is Cauchy-Schwarz.
A: In minimization problems, it always helps to write out the objective function, $(x + m)^2 + (y + n)^2$, which is the precise quantity you want to minimize.  From there you can either use calculus to finish this off rigorously, or play around with it and minimize it manually.  Let me know if neither of those approaches yield results for you.
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 we have 
$$|\langle \vec a,\vec b\rangle|\le ||\vec a||\,||\vec b||\,\tag 2$$
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.
