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?
### Part (A): \( a = 0, b = 1 \)

Given the inequality:
\[
\frac{1}{x(ay + b)} + \frac{1}{y(az + b)} + \frac{1}{z(ax + b)} \geq 3
\]
Substitute \( a = 0 \) and \( b = 1 \):
\[
\frac{1}{x(0 \cdot y + 1)} + \frac{1}{y(0 \cdot z + 1)} + \frac{1}{z(0 \cdot x + 1)} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}
\]
Given \( xyz = 1 \), we have:
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 3
\]
By the AM-GM inequality:
\[
\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq 3 \sqrt[3]{\frac{1}{xyz}} = 3 \sqrt[3]{1} = 3
\]
Thus, the inequality holds.

### Part (B): \( a = 1, b = 0 \)

Given the inequality:
\[
\frac{1}{x(ay + b)} + \frac{1}{y(az + b)} + \frac{1}{z(ax + b)} \geq 3
\]
Substitute \( a = 1 \) and \( b = 0 \):
\[
\frac{1}{x(1 \cdot y + 0)} + \frac{1}{y(1 \cdot z + 0)} + \frac{1}{z(1 \cdot x + 0)} = \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx}
\]
Given \( xyz = 1 \), we have:
\[
\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = z + x + y
\]
By the AM-GM inequality:
\[
x + y + z \geq 3 \sqrt[3]{xyz} = 3 \sqrt[3]{1} = 3
\]
Thus, the inequality holds.

### Part (C): \( a + b = 1, \; a, b > 0 \)

Given the inequality:
\[
\frac{1}{x(ay + b)} + \frac{1}{y(az + b)} + \frac{1}{z(ax + b)} \geq 3
\]
Let \( x = \frac{n}{m}, y = \frac{p}{n}, z = \frac{m}{p} \) with \( m, n, p > 0 \). Then the inequality can be written as:
\[
\frac{m}{ap + bn} + \frac{n}{am + bp} + \frac{p}{an + bm} \geq 3
\]
By the Cauchy-Schwarz inequality:
\[
\left( \frac{m}{ap + bn} + \frac{n}{am + bp} + \frac{p}{an + bm} \right) \left( m(ap + bn) + n(am + bp) + p(an + bm) \right) \geq (m + n + p)^2
\]
Since \( a + b = 1 \), we have:
\[
m(ap + bn) + n(am + bp) + p(an + bm) = (a + b)(mn + np + pm) = mn + np + pm
\]
Thus:
\[
\frac{(m + n + p)^2}{mn + np + pm} \geq 3
\]
By the AM-GM inequality:
\[
m + n + p \geq 3 \sqrt[3]{mnp}
\]
Given \( xyz = 1 \), we have \( mnp = 1 \), so:
\[
m + n + p \geq 3
\]
Thus:
\[
\frac{(m + n + p)^2}{mn + np + pm} \geq \frac{9}{3} = 3
\]
The equality holds if \( x = y = z = 1 \).

The final answer is \(\boxed{3}\)