To solve the inequality \( (x - 1)^2 + x^2 \leq (x + 1)^2 + (x + 2)^2 \), let's first expand both sides.

1. **Expand the left-hand side:**
\[
(x - 1)^2 + x^2 = (x^2 - 2x + 1) + x^2 = 2x^2 - 2x + 1
\]

2. **Expand the right-hand side:**
\[
(x + 1)^2 + (x + 2)^2 = (x^2 + 2x + 1) + (x^2 + 4x + 4) = 2x^2 + 6x + 5
\]

3. **Set up the inequality:**
\[
2x^2 - 2x + 1 \leq 2x^2 + 6x + 5
\]

4. **Subtract \(2x^2\) from both sides:**
\[
-2x + 1 \leq 6x + 5
\]

5. **Rearrange the inequality:**
\[
1 - 5 \leq 6x + 2x
\]
\[
-4 \leq 8x
\]

6. **Divide both sides by 8:**
\[
-\frac{1}{2} \leq x
\]
or equivalently,
\[
x \geq -\frac{1}{2}
\]

**Conclusion:**
The solution to the inequality is \( x \geq -\frac{1}{2} \).