Solution. Denote $x=4(ab+bc+ca)$ and $y=8abc$ then we obtain
$$a^{2}+b^{2}+c^{2}=1-\frac{x}{2} ; \quad a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}=\frac{x^{2}}{16}-\frac{y}{4}$$

We can rewrite the inequality to the form
$$\begin{aligned} & 2 x \geq(4-2 x+8 y)\left(x^{2}-4 y\right) \\ \Leftrightarrow & x(x-1)^{2} \geq 4 y((x-1)(x+2)-4 y) \\ \Leftrightarrow & x(x-1)^{2}+16 y^{2} \geq 4 y(x-1)(x+2) \end{aligned}$$

Obviously, $x \leq \frac{4}{3}$. If $x \leq 1$, we are done immediately. Otherwise, suppose that $x \geq 1$. By the third degree-Schur inequality, it's easy to get $8(x-1) \leq 9 y$. Considering $x$ as a parameter in $\left[1, \frac{4}{3}\right]$, we will prove that $f(y) \geq 0$, where
$$f(y)=16 y^{2}-4 y(x-1)(x+2)+x(x-1)^{2}$$