1. We need the minimum value of $\left|z+\frac12\right|$ given that $|z|\ge 2.$

2. Interpret this geometrically in the Argand plane.

• The condition $|z|\ge 2$ means the point $z$ lies on or outside the circle centered at $0$ with radius $2$.
• The quantity $\left|z+\frac12\right|=\left|z-\left(-\frac12\right)\right|$ is the distance of $z$ from the fixed point $-\frac12$ on the real axis.

So we want the minimum distance from the point $-\frac12$ to the region $|z|\ge 2$.

3. The point $-\frac12$ lies inside the circle $|z|=2$, since $\left|-\frac12\right|=\frac12<2.$ Hence the nearest point in the region $|z|\ge 2$ will be on the boundary circle $|z|=2$, along the line joining the origin and $-\frac12$.

4. Therefore, the minimum distance is $2-\frac12=\frac32.$

We can also verify by triangle inequality: $\left|z+\frac12\right|=\left|z-\left(-\frac12\right)\right|\ge ||z|-\tfrac12|.$ Since $|z|\ge 2$, $\left|z+\frac12\right|\ge 2-\frac12=\frac32.$ Equality occurs when $z$ is on the negative real axis with $|z|=2$, i.e. at $z=-2.$ Then $\left|-2+\frac12\right|=\left|-\frac32\right|=\frac32.$ So the minimum value is exactly $\boxed{\frac32}.$

5. Check options:

• A: strictly greater than $\frac52$ — false
• B: strictly greater than $\frac32$ but less than $\frac52$ — false, because minimum equals $\frac32$, not strictly greater
• C: equal to $\frac52$ — false
• D: lies in interval $(1,2)$ — true, since $\frac32\in(1,2)$

Thus the correct option is $\boxed{D}.$