$|z-3| + |z| + |z+3| = 12$ Let $z$ be a complex number such that $|z-3|+ |z|+ |z+3| = 12$.
If $a = \lfloor|z|\rfloor$ and $b = \lceil|z|\rceil$, where $\lfloor i\rfloor$ denote the greatest integer less than or equal to $i$ and 
$\lceil i\rceil$ denotes the least integer greater or equal to $i$, then find $k = a + b$.
There will be two answers possible.
$z=4$ is a trivial solution, for which value of $k=8$.
There will be one more solution of $z$ for which $k=7$.
I used the triangular inequality:
$$12=|z-3|+|z|+|z+3| \ge |z-3+z+z+3|=3|z|.$$
I used the RMS$\ge$AM inequality to find the upper bound of |z|.
$sqrt{frac{|z-3|^2+|z|^2+|z+3|^2}{3}}\ge \frac{|z-3|+|z|+|z+3|}{3}$
: $|z| \ge \sqrt{10}$
which implies that $a=3$ or $a=4$, $b=4$ hence $k = 7$ or $k=8$.
Suggest some other ways to find the upper limit of $|z|$.

Write $z=r(\cos\phi+i\sin\phi)$. Then we have to investigate the equation
$$\sqrt{r^2-6r\cos\phi+9}+\sqrt{r^2+6r\cos\phi+9}+r=12\ ,$$
which can be written as
$$\sqrt{r^2+9}\ \bigl(\sqrt{1-t\cos\phi}+\sqrt{1+t\cos\phi}\bigr)+r=12\ ,\tag{1}$$
whereby
$$0\leq t:={6r\over r^2+9}\leq1\ .$$
The function
$$s\mapsto\sqrt{1-s}+\sqrt{1+s}=\sqrt{2+2\sqrt{1-s^2}}\qquad(0\leq s\leq1)$$
is monotonically decreasing. It follows that the value of the large parenthesis in $(1)$ is maximal   when $\cos\phi=0$, leading to the condition $2\sqrt{r^2+9}+r=12$. This gives $r_{\min}=\sqrt{52}-4=3.2111$. On the other hand the value of the large parenthesis in $(1)$ is minimal when $\cos\phi=1$, which then leads to $r_{\max}=4$. 
Returning to the original problem we can say that the $z$ fulfilling the given equation have to satisfy
$$3.2111\leq|z|\leq 4\ .$$
This allows of $a=3$ and $b=4$, as well as $a=b=4$. The possible values for $k$ therefore are $7$ and $8$.
