We can start by using the Cauchy-Schwarz inequality:

(a^2 + b^2 + c^2)(1 + 1 + 1) ≥ (a + b + c)^2

Simplifying, we get:

3(a^2 + b^2 + c^2) ≥ (a + b + c)^2

Expanding the denominator on the right-hand side of the inequality we want to prove, we get:

1/(a^2 + b^2 + c^2) + 1/(2(ab + bc + ca))

= (2(ab + bc + ca) + a^2 + b^2 + c^2)/(2(ab + bc + ca)(a^2 + b^2 + c^2))

Multiplying both sides by (a^2 + b^2 + c^2), we get:

2(ab + bc + ca) + a^2 + b^2 + c^2 ≥ 9(ab + bc + ca)

Simplifying, we get:

a^2 + b^2 + c^2 + 2(ab + bc + ca) ≥ 3(ab + bc + ca)

Which is true by the inequality we derived from Cauchy-Schwarz. Therefore, the original inequality is also true.