To solve the inequality \( 4|x^2 - 1| + |x^2 - 4| \geq 6 \), we will analyze the expression by considering the critical points where the expressions inside the absolute values change sign.
### Step 1: Identify Critical Points
The expressions inside the absolute values are \( x^2 - 1 \) and \( x^2 - 4 \). We find the roots:
- \( x^2 - 1 = 0 \) gives \( x = \pm 1 \)
- \( x^2 - 4 = 0 \) gives \( x = \pm 2 \)
Thus, the critical points are \( -2, -1, 1, 2 \). These points divide the number line into intervals:
- \( (-\infty, -2) \)
- \( (-2, -1) \)
- \( (-1, 1) \)
- \( (1, 2) \)
- \( (2, \infty) \)
### Step 2: Analyze Each Interval
**Interval 1: \( (-\infty, -2) \)**
- Here, \( x^2 - 1 \geq 0 \) and \( x^2 - 4 \geq 0 \).
- Therefore, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = x^2 - 4 \).
- The inequality becomes:
\[
4(x^2 - 1) + (x^2 - 4) \geq 6
\]
Simplifying:
\[
4x^2 - 4 + x^2 - 4 \geq 6 \implies 5x^2 - 8 \geq 6 \implies 5x^2 \geq 14 \implies x^2 \geq \frac{14}{5}
\]
Thus, \( x \leq -\sqrt{\frac{14}{5}} \) or \( x \geq \sqrt{\frac{14}{5}} \).
**Interval 2: \( (-2, -1) \)**
- Here, \( x^2 - 1 \geq 0 \) and \( x^2 - 4 < 0 \).
- Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = -(x^2 - 4) = 4 - x^2 \).
- The inequality becomes:
\[
4(x^2 - 1) + (4 - x^2) \geq 6
\]
Simplifying:
\[
4x^2 - 4 + 4 - x^2 \geq 6 \implies 3x^2 \geq 6 \implies x^2 \geq 2 \implies x \leq -\sqrt{2} \text{ or } x \geq \sqrt{2}
\]
However, in this interval, we only consider \( x \leq -1 \).
**Interval 3: \( (-1, 1) \)**
- Here, \( x^2 - 1 < 0 \) and \( x^2 - 4 < 0 \).
- Thus, \( |x^2 - 1| = -(x^2 - 1) = 1 - x^2 \) and \( |x^2 - 4| = 4 - x^2 \).
- The inequality becomes:
\[
4(1 - x^2) + (4 - x^2) \geq 6
\]
Simplifying:
\[
4 - 4x^2 + 4 - x^2 \geq 6 \implies -5x^2 + 8 \geq 6 \implies -5x^2 \geq -2 \implies x^2 \leq \frac{2}{5}
\]
Thus, \( -\sqrt{\frac{2}{5}} \leq x \leq \sqrt{\frac{2}{5}} \).
**Interval 4: \( (1, 2) \)**
- Here, \( x^2 - 1 \geq 0 \) and \( x^2 - 4 < 0 \).
- Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = 4 - x^2 \).
- The inequality becomes:
\[
4(x^2 - 1) + (4 - x^2) \geq 6
\]
Simplifying:
\[
4x^2 - 4 + 4 - x^2 \geq 6 \implies 3x^2 \geq 6 \implies x^2 \geq 2 \implies x \geq \sqrt{2}
\]
In this interval, we consider \( 1 < x < 2 \).
**Interval 5: \( (2, \infty) \)**
- Here, \( x^2 - 1 \geq 0 \) and \( x^2 - 4 \geq 0 \).
- Thus, \( |x^2 - 1| = x^2 - 1 \) and \( |x^2 - 4| = x^2 - 4 \).
- The inequality becomes:
\[
4(x^2 - 1) + (x^2 - 4) \geq 6
\]
Simplifying:
\[
4x^2 - 4 + x^2 - 4 \geq 6 \implies 5x^2 - 8 \geq 6 \implies 5x^2 \geq 14 \implies x^2 \geq \frac{14}{5}
\]
Thus, \( x \geq \sqrt{\frac{14}{5}} \).
### Step 3: Combine Results
Now we combine the results from each interval:
1. From \( (-\infty, -2) \): \( x \leq -\sqrt{\frac{14}{5}} \)
2. From \( (-2, -1) \): \( x \leq -\sqrt{2} \)
3. From \( (-1, 1) \): \( -\sqrt{\frac{2}{5}} \leq x \leq \sqrt{\frac{2}{5}} \)
4. From \( (1, 2) \): \( x \geq \sqrt{2} \)
5. From \( (2, \infty) \): \( x \geq \sqrt{\frac{14}{5}} \)
### Final Solution
The final solution set is:
\[
x \in (-\infty, -\sqrt{\frac{14}{5}}] \cup [-\sqrt{2}, \sqrt{2}] \cup [\sqrt{2}, \infty)
\]
