To prove that \(a^2 + b^2 + c^2 \geq 6\), given that \(a + b + c + ab + bc + ca = 6\), where \(a\), \(b\), and \(c\) are positive real numbers, we can use the identity for the square of a sum:

\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)
\]

From the given equation \(a + b + c + ab + bc + ca = 6\), we have:

Let \(x = a + b + c\) and \(y = ab + bc + ca\).

So \(x + y = 6\).

We need to find \(a^2 + b^2 + c^2\):

Using the identity, we rearrange:

\[
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca)
\]

Substitute \(x\) and \(y\):

\[
a^2 + b^2 + c^2 = x^2 - 2y
\]

We know:

\[
x + y = 6 \quad \Rightarrow \quad y = 6 - x
\]

Substitute \(y\) into the identity for \(a^2 + b^2 + c^2\):

\[
a^2 + b^2 + c^2 = x^2 - 2(6 - x) = x^2 - 12 + 2x
\]

Thus, we want:

\[
x^2 + 2x - 12 \geq 6
\]

This simplifies to:

\[
x^2 + 2x - 18 \geq 0
\]

To solve this quadratic inequality, first find the roots of the equation:

\[
x^2 + 2x - 18 = 0
\]

Using the quadratic formula:

\[
x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-18)}}{2 \cdot 1}
\]
\[
x = \frac{-2 \pm \sqrt{4 + 72}}{2}
\]
\[
x = \frac{-2 \pm \sqrt{76}}{2}
\]
\[
x = \frac{-2 \pm 2\sqrt{19}}{2}
\]
\[
x = -1 \pm \sqrt{19}
\]

Hence, the roots are \(x = -1 + \sqrt{19}\) and \(x = -1 - \sqrt{19}\).

Since \(x = a + b + c\) is positive, only \(x = -1 + \sqrt{19}\) is relevant for the inequality \(x^2 + 2x - 18 \geq 0\).

In conclusion, \(a^2 + b^2 + c^2 \geq 6\) for positive \(a\), \(b\), and \(c\) satisfying the given condition.