Stable gain nyquist plot From Modern Control Engineering 5th Edition page 469

The transfer function of a plant controller by a proportional controller is given by 

$$G(s) = \frac{K(s+0.5)}{s^3 + s^2 + 1}$$

In the book $G(j\omega) / K$ is plotted and from that they infer that for $ 2 &lt; K$ the system is stable.

I plotted the nyquist plot of the plant itself



and tried getting the answer consistent with the procedure from our lecture.

I know that I need 2 counter clockwise encirclements of $(-1,0j)$ so I want to see for which gain the plot is shifted past -1.

$$\frac{-1}{K} &gt; -0.5$$

$$K &gt; 2$$

So far so good. They did not compute the upper limit of $K$.

I think I can see that the encirclements occur between -0.5 and 0 from the plot. 

So next I need to calculate 

$$\frac{-1}{K} &lt; 0$$

$$\frac{1}{K} &gt; 0$$

$$K &gt; \infty$$

and this can't be. I assumed to get to $$K &lt; \infty$$

I think my main problem is that I don't exactly know when to use &lt; and > and which points (-0.5,0,0.5) I should choose for my calculations. Also how would I proceed if there was only 1 RHP and I'd need only 1 ccw encirclement.

Could someone clarify the general procedure for this or guide me to a ressource that explains it for beginners?
There is a mistake in the last step. When you invert both sides of the inequality, the $&gt;$ inverts as well. So, for $K&gt;0$ (as you found that $K&gt;2$, so it must be positive):

$$\frac 1 K &gt; y \iff K&lt;\frac 1 y$$

If you do this when $y\to 0$, then you get the desired result.

Also, note that $\frac 1 K &gt;0$ would hold for any $K&gt;0$, as the only way to get a negative number from $1/K$ is letting $K$ be negative.

Regarding your other question, it would completely depend on the transfer function and its respective Nyquist. But don't worry, the procedure is exactly the same. For example, try doing the Nyquist plot for $-\frac{K}{1-s}$ and see what you get.