Let $\displaystyle {x, y, z}$ be positive real numbers such that $\displaystyle {xyz = 1}$. Prove the inequality:$$\displaystyle{\dfrac{1}{x\left(ay+b\right)}+\dfrac{1}{y\left(az+b\right)}+\dfrac{1}{z\left(ax+b\right)}\geq 3}$$
if:
(A) $\displaystyle {a = 0, b = 1}$
(B) $\displaystyle {a = 1, b = 0}$
(C) $\displaystyle {a + b = 1, \; a, b> 0}$
When the equality holds?