Investigating the domain of definition of functions.
(a) Solve the inequality `sqrt(x+2) +sqrt(x-5) ge sqrt(5-x)`
(b) Prove that the equality
`log_(2-x) (x-3) ge -5` has no solution.

Answer

Let's solve the given problems step by step.

### Part (a): Solve the inequality

**Inequality:**
\[ \sqrt{x+2} + \sqrt{x-5} \geq \sqrt{5-x} \]

**Step 1: Determine the domain of the functions involved.**
For the square roots to be defined, we need:
1. \(x + 2 \geq 0 \Rightarrow x \geq -2\)
2. \(x - 5 \geq 0 \Rightarrow x \geq 5\)
3. \(5 - x \geq 0 \Rightarrow x \leq 5\)

Combining these conditions, we find that \(x\) must satisfy:
\[ x \geq 5 \quad \text{and} \quad x \leq 5 \]
Thus, the only value that satisfies these conditions is:
\[ x = 5 \]

**Step 2: Substitute \(x = 5\) into the inequality to check if it holds.**
Substituting \(x = 5\):
\[ \sqrt{5 + 2} + \sqrt{5 - 5} \geq \sqrt{5 - 5} \]
This simplifies to:
\[ \sqrt{7} + 0 \geq 0 \]
This is true since \(\sqrt{7} > 0\).

**Conclusion for part (a):**
The solution to the inequality is:
\[ \{5\} \]

### Part (b): Prove that the equality

**Inequality:**
\[ \log_{(2-x)}(x-3) \geq -5 \]

**Step 1: Determine the conditions for the logarithm to be defined.**
For the logarithm to be defined, we need:
1. \(2 - x > 0 \Rightarrow x < 2\)
2. \(x - 3 > 0 \Rightarrow x > 3\)

**Step 2: Analyze the inequalities.**
From \(x < 2\) and \(x > 3\), there is no \(x\) that can satisfy both conditions simultaneously. Thus, there is no solution.

**Conclusion for part (b):**
The equality \(\log_{(2-x)}(x-3) \geq -5\) has no solution.

### Summary of Solutions:
- Part (a): The solution is \(x = 5\).
- Part (b): There is no solution.