Certainly! Let's solve the given inequality step by step:
[tex]\[
(2x + 1)^2 + 3(x + 2)^2 \geq \left(x + \frac{1}{2}\right)^2
\][/tex]
### Step 1: Expand each squared term
Expand [tex]\((2x + 1)^2\)[/tex]:
[tex]\[
(2x + 1)^2 = (2x)^2 + 2 \cdot 2x \cdot 1 + 1^2 = 4x^2 + 4x + 1
\][/tex]
Expand [tex]\(3(x + 2)^2\)[/tex]:
[tex]\[
3(x + 2)^2 = 3[(x)^2 + 2 \cdot x \cdot 2 + 2^2] = 3(x^2 + 4x + 4) = 3x^2 + 12x + 12
\][/tex]
Expand [tex]\(\left(x + \frac{1}{2}\right)^2\)[/tex]:
[tex]\[
\left(x + \frac{1}{2}\right)^2 = x^2 + 2 \cdot x \cdot \frac{1}{2} + \left(\frac{1}{2}\right)^2 = x^2 + x + \frac{1}{4}
\][/tex]
### Step 2: Substitute the expanded terms into the inequality
Now, replace the expanded forms back into the original inequality:
[tex]\[
(4x^2 + 4x + 1) + (3x^2 + 12x + 12) \geq (x^2 + x + \frac{1}{4})
\][/tex]
Combine like terms on the left side:
[tex]\[
4x^2 + 3x^2 + 4x + 12x + 1 + 12 \geq x^2 + x + \frac{1}{4}
\][/tex]
[tex]\[
7x^2 + 16x + 13 \geq x^2 + x + \frac{1}{4}
\][/tex]
### Step 3: Bring all terms to one side of the inequality
Subtract [tex]\(x^2 + x + \frac{1}{4}\)[/tex] from both sides:
[tex]\[
7x^2 + 16x + 13 - x^2 - x - \frac{1}{4} \geq 0
\][/tex]
Combine like terms:
[tex]\[
6x^2 + 15x + \left(13 - \frac{1}{4}\right) \geq 0
\][/tex]
[tex]\[
6x^2 + 15x + \frac{51}{4} \geq 0
\][/tex]
### Conclusion
The derived inequality is:
[tex]\[
6x^2 + 15x + \frac{51}{4} \geq 0
\][/tex]
Given the solution, this inequality is satisfied for all real numbers [tex]\(x \in \mathbb{R}\)[/tex]. Therefore, the solution to the inequality is:
[tex]\[
\boxed{x \in (-\infty, \infty)}
\][/tex]
