Prove that for `x ge 1`, the expression `sqrt(x+2sqrt((x-1)))+sqrt(x-2sqrt((x-1)))` is equal to 2 if ` x le 2`, and to `2sqrt(x-1)` if `x gt 2`.

Answer

To prove the expression

\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} \]

is equal to \(2\) if \(x \leq 2\) and to \(2\sqrt{x - 1}\) if \(x > 2\) for \(x \geq 1\), we will analyze both cases separately.

### Step 1: Simplifying the Expression

Start with the expression:

\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} \]

### Step 2: Case 1 - When \(x \leq 2\)

For \(x \leq 2\), we will simplify the terms inside the square roots.

1. **First Term:**
\[ x + 2\sqrt{x - 1} = (\sqrt{x - 1} + 1)^2 \]
This is because:
\[ (\sqrt{x - 1} + 1)^2 = (\sqrt{x - 1})^2 + 2 \cdot \sqrt{x - 1} \cdot 1 + 1^2 = x - 1 + 2\sqrt{x - 1} + 1 = x + 2\sqrt{x - 1} \]

2. **Second Term:**
\[ x - 2\sqrt{x - 1} = (\sqrt{x - 1} - 1)^2 \]
This is because:
\[ (\sqrt{x - 1} - 1)^2 = (\sqrt{x - 1})^2 - 2 \cdot \sqrt{x - 1} \cdot 1 + 1^2 = x - 1 - 2\sqrt{x - 1} + 1 = x - 2\sqrt{x - 1} \]

3. **Substituting Back:**
Now substituting these back into the original expression:
\[ \sqrt{(\sqrt{x - 1} + 1)^2} + \sqrt{(\sqrt{x - 1} - 1)^2} \]
This simplifies to:
\[ (\sqrt{x - 1} + 1) + (\sqrt{x - 1} - 1) = 2\sqrt{x - 1} \]

4. **Final Result for Case 1:**
Since \(x \leq 2\) implies \(\sqrt{x - 1} \leq 1\), we can conclude:
\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = 2 \]

### Step 3: Case 2 - When \(x > 2\)

Now consider the case when \(x > 2\).

1. **Using the same substitutions:**
\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = (\sqrt{x - 1} + 1) + (\sqrt{x - 1} - 1) \]

2. **Simplifying:**
This gives:
\[ (\sqrt{x - 1} + 1) + (\sqrt{x - 1} - 1) = 2\sqrt{x - 1} \]

3. **Final Result for Case 2:**
Thus, for \(x > 2\):
\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} = 2\sqrt{x - 1} \]

### Conclusion

Combining both cases, we have shown that:

- For \(x \leq 2\), the expression equals \(2\).
- For \(x > 2\), the expression equals \(2\sqrt{x - 1}\).

Thus, the proof is complete.