Solution:
By using the Arithmetic-Geometric Mean Inequality for 15 positive numbers, we find that
$$\begin{aligned} & a^{2} + b^{2} + c^{2} + a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} = \\ & \quad = \frac{a^{2}}{4} + \frac{a^{2}}{4} + \frac{a^{2}}{4} + \frac{a^{2}}{4} + \frac{b^{2}}{4} + \frac{b^{2}}{4} + \frac{b^{2}}{4} + \frac{b^{2}}{4} + \frac{c^{2}}{4} + \frac{c^{2}}{4} + \frac{c^{2}}{4} + \frac{c^{2}}{4} + a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} \geq \\ & \quad \geq 15 \sqrt[15]{\frac{a^{12} b^{12} c^{12}}{4^{12}}} = 15 \sqrt[15]{\left(\frac{a b c}{4}\right)^{12}} = 15 \sqrt[15]{\left(\frac{1}{32}\right)^{12}} = \frac{15}{16} \end{aligned}$$
as desired. Equality holds if and only if $a = b = c = \frac{1}{2}$.

By using AM-GM we obtain
$$\begin{aligned} & \left(a^{2} + b^{2} + c^{2}\right) + \left(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}\right) \geq 3 \sqrt[3]{a^{2} b^{2} c^{2}} + 3 \sqrt[3]{a^{4} b^{4} c^{4}} = \\ & = 3 \sqrt[3]{\left(\frac{1}{8}\right)^{2}} + 3 \sqrt[3]{\left(\frac{1}{8}\right)^{4}} = \frac{3}{4} + \frac{3}{16} = \frac{15}{16} \end{aligned}$$
The equality holds when $a^{2} = b^{2} = c^{2}$, i.e. $a = b = c = \frac{1}{2}$.