Since $a^2 + b^2 + c^2 = 3$, the inequality is equivalent to

$$\left( \frac{a^2 + b^2}{2ab} + 1 \right) + \left( \frac{b^2 + c^2}{2bc} + 1 \right) + \left( \frac{c^2 + a^2}{2ca} + 1 \right) + \left( \frac{2(ab + bc + ca)}{a^2 + b^2 + c^2} + 1 \right) \geq 9.$$
which can be written as
$$\frac{(a+b)^2}{2ab} + \frac{(b+c)^2}{2bc} + \frac{(c+a)^2}{2ca} + \frac{(a+b+c)^2}{a^2+b^2+c^2} \geq 9.$$
Let $L$ be the left hand side of the last inequality. Then from Cauchy-Schwarz inequality in the form of Engel or Andreescu it follows that
$$L \geq \frac{((a+b) + (b+c) + (c+a) + (a+b+c))^2}{2ab + 2bc + 2ca + a^2 + b^2 + c^2} = \frac{9(a+b+c)^2}{(a+b+c)^2} = 9.$$
The equality holds, if and only if
$$\frac{a+b}{2ab} = \frac{b+c}{2bc} = \frac{c+a}{2ca} = \frac{a+b+c}{3}.$$
Since $a, b, c > 0$, from the first two equations we get $a = b = c$, and by using the last equation we find: $a = b = c = 1$.