What is the minimum value of $|z+\frac{1}{2}|$ for $|z|\geq 2$ Find the minimum value of $|z+\frac{1}{2}|$ for $|z|\geq 2$.    
Is the following approaches correct ?   
$\left|z+\frac{1}{2} \right|=\sqrt{(x+\frac{1}{2})^2+y^2}=\sqrt{x^2+y^2+x+\frac{1}{4}}$. What will be the next?   
Update: 
$|z|\geq 2 \implies x^2+y^2\geq 4\implies x^2+y^2+x+\frac{1}{4}\geq 4+\frac{1}{4}+x\implies \sqrt{x^2+y^2+x+\frac{1}{4}} \geq \sqrt{\frac{17}{4}+x}$. 

Use the triangle inequality:
$$||a|-|b|| \leq |a+b| \leq |a|+|b|.$$
You need only the first part with $a=z$ and $b=1/2$. So
$$ |z+\frac12| \geq ||z|-\frac12| = |z| - \frac12 \geq 2 - 0.5 = 1.5$$
It is enough to notice, that for $z=-2$ we have $|z|=2$ and $|z+\frac12| = 1.5$.
More about triangle inequality:
https://en.wikipedia.org/wiki/Triangle_inequality
