To prove that the quadratic expression \( x^2 + x + 1 \) is always greater than 0 for all real numbers \( x \), we can analyze its properties.

1. **Discriminant Analysis**:
The discriminant of a quadratic equation \( ax^2 + bx + c \) is given by \( \Delta = b^2 - 4ac \). For the quadratic \( x^2 + x + 1 \), we have:
\[
a = 1, \quad b = 1, \quad c = 1
\]
The discriminant is:
\[
\Delta = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3
\]
Since the discriminant is negative (\( \Delta < 0 \)), the quadratic equation \( x^2 + x + 1 = 0 \) has no real roots. This implies that the quadratic expression does not cross the x-axis and is either always positive or always negative.

2. **Leading Coefficient**:
The leading coefficient \( a \) of the quadratic \( x^2 + x + 1 \) is positive (\( a = 1 \)). For a quadratic expression with a positive leading coefficient and no real roots, the parabola opens upwards and the expression is always positive.

3. **Vertex Analysis**:
The vertex form of a quadratic \( ax^2 + bx + c \) can be found using the vertex formula \( x = -\frac{b}{2a} \). For \( x^2 + x + 1 \):
\[
x = -\frac{b}{2a} = -\frac{1}{2 \cdot 1} = -\frac{1}{2}
\]
Substituting \( x = -\frac{1}{2} \) back into the quadratic to find the minimum value:
\[
f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4}
\]
The minimum value of the quadratic expression \( x^2 + x + 1 \) is \( \frac{3}{4} \), which is greater than 0.

Since the quadratic expression \( x^2 + x + 1 \) has no real roots, opens upwards, and has a minimum value of \( \frac{3}{4} \), we can conclude that:
\[
x^2 + x + 1 > 0 \quad \text{for all real numbers } x.
\]