To prove the inequality $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq a + b + c$ for positive $a, b, c$, we will use the given inequality $\frac{3}{abc} \geq a + b + c$.

1. Given Inequality:
$$\frac{3}{abc} \geq a + b + c$$

2. Rewrite the Given Inequality:
Multiply both sides by $abc$:
$$3 \geq abc(a + b + c)$$

3. **Assume $a, b, c$ are positive:**
Since $a, b, c > 0$, we can divide both sides by $abc$:
$$\frac{3}{abc} \geq a + b + c$$

4. Use the AM-HM Inequality:
The Arithmetic Mean-Harmonic Mean (AM-HM) inequality states that for positive numbers $a, b, c$:
$$\frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

5. Rearrange the AM-HM Inequality:
Multiply both sides by 3:
$$a + b + c \geq \frac{9}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$

6. Invert the Inequality:
Taking the reciprocal of both sides (since all terms are positive):
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{9}{a + b + c}$$

7. Combine with the Given Inequality:
From step 2, we have:
$$3 \geq abc(a + b + c)$$
Since $abc > 0$, we can divide both sides by $abc$:
$$\frac{3}{abc} \geq a + b + c$$

8. Conclusion:
Combining the results from steps 6 and 7, we get:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq a + b + c$$

Thus, we have proven the required inequality.

$\blacksquare$

The final answer is $\boxed{ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq a + b + c }$