1. Let the radii of the three circles be $x, y,$ and $z$. Since the circles are externally tangent, the length of the common external tangent between any two circles is given by the sum of their radii. Therefore, we have:
$$a = x + y, \quad b = y + z, \quad c = z + x$$

2. We need to show that:
$$\pi (a + b + c)^2 \le 12T$$
where $T$ is the sum of the areas of the circles. The area of a circle with radius $r$ is $\pi r^2$. Thus, the total area $T$ is:
$$T = \pi x^2 + \pi y^2 + \pi z^2$$

3. First, calculate $a + b + c$:
$$a + b + c = (x + y) + (y + z) + (z + x) = 2x + 2y + 2z = 2(x + y + z)$$

4. Now, square $a + b + c$:
$$(a + b + c)^2 = (2(x + y + z))^2 = 4(x + y + z)^2$$

5. We need to show that:
$$\pi (a + b + c)^2 \le 12T$$
Substituting the expression for $(a + b + c)^2$ and $T$, we get:
$$\pi \cdot 4(x + y + z)^2 \le 12 (\pi x^2 + \pi y^2 + \pi z^2)$$

6. Simplify the inequality:
$$4(x + y + z)^2 \le 12 (x^2 + y^2 + z^2)$$

7. Divide both sides by 4:
$$(x + y + z)^2 \le 3 (x^2 + y^2 + z^2)$$

8. The inequality $(x + y + z)^2 \le 3 (x^2 + y^2 + z^2)$ is a direct application of the Cauchy-Schwarz inequality in the form:
$$(x^2 + y^2 + z^2)(1^2 + 1^2 + 1^2) \ge (x + y + z)^2$$
Simplifying the right-hand side, we get:
$$3(x^2 + y^2 + z^2) \ge (x + y + z)^2$$
which is exactly what we needed to prove.

Therefore, the inequality holds true.

$\blacksquare$