Complete set of solution of inequation `sqrt(x+2sqrt(x-1))+sqrt(x-2sqrt(x-1))>3/2`

Options

A. `(-oo,1)`
B. `(-oo,1]`
C. `[1,oo)`
D. `(1,oo)`

Answer

To solve the inequality \( \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} > \frac{3}{2} \), we will follow these steps:

### Step 1: Determine the domain of the expressions
The term \( \sqrt{x - 1} \) is defined when \( x - 1 \geq 0 \), which gives us:
\[ x \geq 1 \]
Additionally, we need to ensure that \( \sqrt{x - 2\sqrt{x - 1}} \) is also defined. This term is defined when:
\[ x - 2\sqrt{x - 1} \geq 0 \]
Rearranging gives:
\[ x \geq 2\sqrt{x - 1} \]
Squaring both sides (noting that both sides are non-negative for \( x \geq 1 \)):
\[ x^2 \geq 4(x - 1) \]
This simplifies to:
\[ x^2 - 4x + 4 \geq 0 \]
Factoring gives:
\[ (x - 2)^2 \geq 0 \]
This inequality is always true for all \( x \). Therefore, the only restriction on \( x \) is \( x \geq 1 \).

### Step 2: Analyze the inequality
We need to solve:
\[ \sqrt{x + 2\sqrt{x - 1}} + \sqrt{x - 2\sqrt{x - 1}} > \frac{3}{2} \]

### Step 3: Simplify the terms
Let \( \sqrt{x - 1} = t \), then \( x = t^2 + 1 \). Substituting this into the inequality gives:
\[ \sqrt{(t^2 + 1) + 2t} + \sqrt{(t^2 + 1) - 2t} > \frac{3}{2} \]
This simplifies to:
\[ \sqrt{(t + 1)^2} + \sqrt{(t - 1)^2} > \frac{3}{2} \]
Which further simplifies to:
\[ |t + 1| + |t - 1| > \frac{3}{2} \]

### Step 4: Analyze the absolute values
We consider two cases based on the value of \( t \):

**Case 1:** \( t \geq 1 \) (i.e., \( \sqrt{x - 1} \geq 1 \))
\[ (t + 1) + (t - 1) > \frac{3}{2} \implies 2t > \frac{3}{2} \implies t > \frac{3}{4} \]
Since \( t \geq 1 \), this condition is satisfied for all \( t \geq 1 \).

**Case 2:** \( t < 1 \) (i.e., \( \sqrt{x - 1} < 1 \))
In this case, we have:
\[ -(t + 1) + (1 - t) > \frac{3}{2} \implies -2t > \frac{3}{2} \implies t < -\frac{3}{4} \]
This case does not provide any valid solutions since \( t \) must be non-negative.

### Step 5: Conclusion
From the analysis, the only valid solutions come from the first case where \( t \geq 1 \), which corresponds to:
\[ \sqrt{x - 1} \geq 1 \implies x - 1 \geq 1 \implies x \geq 2 \]

Thus, the complete set of solutions for the inequality is:
\[ \boxed{[2, \infty)} \]