How to show that the delay margin is zero if the open loop gain $|L(i\infty)| \geq 1$ ?
Where $L(s)$ is the open loop transfer function and the delay margin is the amount of time delay for the system to be on the verge of instability.
Hint: Using the Nyquist Contour.
Nyquist's stability criterion requires that the number of times $L(i\omega)$ encircles $(-1|0)$ equals the number of poles in the open right half plane.
For a non-vanishing delay $T_d \neq 0$, $L(i\omega)e^{-i\omega T_d}$ encircles $(-1|0)$ infinitely often since $\lim\limits_{\omega \rightarrow \infty}|L(i\omega)| \ge 1$, by proposition. Given a finite number of poles we fail to fulfill the stability criterion.