Find minimum value of $|2z-1|+|3z-2|;\,\,z\in\mathbb{C}$? I tried this question using many different ways (triangle inequality, geometric interpretation, etc) but I didn't get the correct answer.

The minimum value of $|2z-1|+|3z-2|;\,\,z\in\mathbb{C}$ is:
  (a) 0
  (b) 1/2
  (c) 1/3
  (d) 2/3

The given answer is 1/3.
The explanation in the book is not clear. Can someone please explain this?
The explanation in the book is:

$\frac12=|−\frac12|=|(3z−2)−(3/2)(2z−1)|$
  $\leq|3z−2|+\frac32|2z−1|$
  $\implies \frac12 \cdot \frac23≤\frac23|3z−2|+|2z−1|$
  So Minimum value is $\frac13$.

 A: First observe that the minimum will never be achieved if $z$ has an imaginary component.  Since $|2(a+bi)-1| = \sqrt{(2a-1)^2 + (2b)^2} > |2a-1|$.  So you can limit your region of interest to the real numbers.  
After that the function (of one variable) is piecewise linear, so the minimum will have to occur at one of the nodes.  Check $z=\tfrac 12$ and $z = \tfrac 23$, and the answer comes out.
The book explanation shows that that the minimum is at most $\tfrac 13$, but, as far as I can tell, doesn't show that it's in fact achieved.
A: You see that the last inequality in the book explanation is
$$
|2z-1|+|3z-2| \ge \frac 1 3
$$
which tells you that the minimum is not less that $\frac 1 3$. On the other hand in any point where the triangle inequality is actually an equality (e.g. for $z= 1/2$) you get an equality. Hence $\frac 1 3$ is the minimum value.
A: Using $|z_1|+|z_2|\geq|z_1+z_2|$
$$
|2z-1|+|3z-2|=2\Big|z-\frac{1}{2}\Big|+3\Big|z-\frac{2}{3}\Big|=2\Big|z-\frac{1}{2}\Big|+2\Big|z-\frac{2}{3}\Big|+\Big|z-\frac{2}{3}\Big|\\
=2\bigg(\Big|z-\frac{1}{2}\Big|+\Big|\frac{2}{3}-z\Big|\bigg)+\Big|z-\frac{2}{3}\Big|\geq2\Big|\frac{2}{3}-\frac{1}{2}\Big|+\Big|z-\frac{2}{3}\Big|\\
\geq2.\frac{1}{6}+\Big|z-\frac{2}{3}\Big|=\frac{1}{3}+\Big|z-\frac{2}{3}\Big|\geq\frac{1}{3}
$$
