To solve the inequality $\left| x + 1 \right|^2 + 2\left| x + 2 \right| \geq 2$, we'll break it down into cases based on the expressions inside the absolute values.

Step 1: Simplify the inequality

First, simplify the inequality by subtracting 2 from both sides:

$\left| x + 1 \right|^2 + 2\left| x + 2 \right| - 2 \geq 0$

$\left| x + 1 \right|^2 \geq 2 - 2\left| x + 2 \right|$

Step 2: Consider different cases

Case 1: $x \geq -1$

• $\left| x + 1 \right| = x + 1$
• $\left| x + 2 \right| = x + 2$

The inequality becomes:

$(x + 1)^2 + 2(x + 2) \geq 2$

Expanding and simplifying:

$(x^2 + 2x + 1) + 2x + 4 \geq 2$

$x^2 + 4x + 5 \geq 2$

$x^2 + 4x + 3 \geq 0$

Factoring:

$(x + 3)(x + 1) \geq 0$

The roots are $x = -3$ and $x = -1$. The inequality is satisfied when $x \leq -3$ or $x \geq -1$.

Since $x \geq -1$ was our initial assumption, the solution in this case is $x \geq -1$.

Case 2: $x < -1$ and $x \geq -2$

• $\left| x + 1 \right| = -(x + 1)$
• $\left| x + 2 \right| = x + 2$

The inequality becomes:

$(-(x + 1))^2 + 2(x + 2) \geq 2$

Expanding and simplifying:

$(x + 1)^2 + 2x + 4 \geq 2$

This results in the same inequality as in Case 1:

$x^2 + 4x + 3 \geq 0$

Given $x < -1$ and $x \geq -2$, the solution in this case is $x \geq -2$ because the inequality is satisfied for $x \leq -3$ or $x \geq -1$.

Case 3: $x < -2$

• $\left| x + 1 \right| = -(x + 1)$
• $\left| x + 2 \right| = -(x + 2)$

The inequality becomes:

$(x + 1)^2 - 2(x + 2) \geq 2$

Expanding and simplifying:

$(x^2 + 2x + 1) - 2x - 4 \geq 2$

$x^2 - 3 \geq 2$

$x^2 \geq 5$

$x \leq -\sqrt{5} \quad \text{or} \quad x \geq \sqrt{5}$

Since $x < -2$, the solution in this case is $x \leq -\sqrt{5}$.

Final Solution

Combine the results from all cases:

$x \leq -\sqrt{5} \quad \text{or} \quad x \geq -1$

Thus, the solution to the inequality $\left| x + 1 \right|^2 + 2\left| x + 2 \right| \geq 2$ is:

$x \in (-\infty, -\sqrt{5}] \cup [-1, \infty)$

Would you like further details on any step, or do you have any questions?

Here are five related questions to expand your understanding:

1. What would happen if the inequality were strictly greater than, rather than greater than or equal to?
2. How would the solution change if the absolute value expressions involved were inside a square root?
3. Can you solve $\left| x + 3 \right| + \left| x - 2 \right| \geq 5$ using similar methods?
4. What is the impact on the solution set if one of the absolute value expressions is replaced with a polynomial?
5. How would this inequality behave if instead of 2, we had a different constant on the right side?

Tip: When dealing with inequalities involving absolute values, always consider breaking the problem into different cases depending on the critical points that make the expressions inside the absolute values zero.