Proof: First, we prove that if $x \geq 0, y \geq 0, x+y \leq 9$, then the inequality holds:
$$\sqrt{9-x}+\sqrt{9-y} \geq 3+\sqrt{9-x-y}$$

Indeed,
$$\begin{aligned} \sqrt{9-x}+\sqrt{9-y} & =\sqrt{(\sqrt{9-x}+\sqrt{9-y})^{2}} \\ & =\sqrt{18-x-y+2 \sqrt{9-x} \sqrt{9-y}} \\ & =\sqrt{18-x-y+2 \sqrt{9^{2}-9 x-9 y+x y}} \\ & \geq \sqrt{18-x-y+6 \sqrt{9-x-y}} \\ & =\sqrt{(3+\sqrt{9-x-y})^{2}} \\ & =3+\sqrt{9-x-y} \end{aligned}$$

Notice that, by $a, b, c \geq 0$, we have $ab + bc + ca \geq 0$, so $a^2 + b^2 + c^2 = 1 - 2(ab + bc + ca) \leq 1$. Therefore, applying inequality $(*)$, we get
$$\begin{aligned} & \sqrt{9-8a^2}+\sqrt{9-8b^2}+\sqrt{9-8c^2} \\ \geq & 3+\sqrt{9-8a^2-8b^2}+\sqrt{9-8c^2} \\ \geq & 3+3+\sqrt{9-8a^2-8b^2-8c^2} \\ = & 6+\sqrt{9-8(a^2+b^2+c^2)} \\ \geq & 7 \end{aligned}$$

Thus, $\sqrt{9-8a^2}+\sqrt{9-8b^2}+\sqrt{9-8c^2} \geq 7$.
Note: The proof of inequality $(*)$ can also be done using derivatives.