Let $a,b\in \mathbb{R}$ and $z\in \mathbb{C}\backslash \mathbb{R}$ so that $\left| a-b \right|=\left| a+b-2z \right|$.
a)	Prove that the equation ${{\left| z-a \right|}^{x}}+{{\left| \bar{z}-b \right|}^{x}}={{\left| a-b \right|}^{x}}$, with the unknown number $x\in \mathbb{R}$, has a unique solution.
b)	Solve the following inequation ${{\left| z-a \right|}^{x}}+{{\left| \bar{z}-b \right|}^{x}}\le {{\left| a-b \right|}^{x}}$, with the unknown number $x\in \mathbb{R}$.
The Mathematical Gazette
### Part (a)

1. **Given Condition:**
   \[
   |a - b| = |a + b - 2z|
   \]
   This implies that the distance between \(a\) and \(b\) is equal to the distance between \(a + b\) and \(2z\).

2. **Equation to Prove:**
   \[
   |z - a|^x + |\bar{z} - b|^x = |a - b|^x
   \]

3. **Using the Given Condition:**
   \[
   |a - b| = |a + b - 2z|
   \]
   Let \(d = |a - b|\). Then:
   \[
   d = |a + b - 2z|
   \]

4. **Expressing \(z\) in Terms of \(a\) and \(b\):**
   Since \(z \in \mathbb{C} \setminus \mathbb{R}\), we can write \(z = x + yi\) where \(y \neq 0\). 

5. **Using the Triangle Inequality:**
   \[
   |z - a| + |z - b| \geq |a - b|
   \]
   Since \(z\) is not real, the equality holds only if \(z\) lies on the line segment joining \(a\) and \(b\), which is not possible here.

6. **Simplifying the Equation:**
   \[
   |z - a|^x + |\bar{z} - b|^x = |a - b|^x
   \]
   Let \(f(x) = |z - a|^x + |\bar{z} - b|^x - |a - b|^x\).

7. **Analyzing \(f(x)\):**
   \[
   f(x) = 0 \implies |z - a|^x + |\bar{z} - b|^x = |a - b|^x
   \]
   We need to show that \(f(x) = 0\) has a unique solution.

8. **Checking \(x = 2\):**
   \[
   |z - a|^2 + |\bar{z} - b|^2 = |a - b|^2
   \]
   This is true because:
   \[
   |z - a|^2 + |\bar{z} - b|^2 = |z - a|^2 + |z - b|^2 = |a - b|^2
   \]
   Hence, \(x = 2\) is a solution.

9. **Uniqueness:**
   Since \(f(x)\) is strictly decreasing, there can be only one solution. Therefore, \(x = 2\) is the unique solution.

\(\blacksquare\)

### Part (b)

1. **Given Inequation:**
   \[
   |z - a|^x + |\bar{z} - b|^x \leq |a - b|^x
   \]

2. **Analyzing the Function:**
   From part (a), we know that \(f(x) = |z - a|^x + |\bar{z} - b|^x - |a - b|^x\) is strictly decreasing.

3. **Behavior of \(f(x)\):**
   Since \(f(x)\) is strictly decreasing and \(f(2) = 0\), for \(x \geq 2\), \(f(x) \leq 0\).

4. **Conclusion:**
   \[
   x \geq 2
   \]

\(\blacksquare\)

The final answer is \( \boxed{ x \geq 2 } \)