Solution:

It is easy to observe that there is a triangle with sides $a+\frac{b}{2}$, $b+\frac{c}{2}$, $c+\frac{a}{2}$. Using Heron's formula, we get
$$16[ABC]^2 = (a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
and
$$16\left[A_1 B_1 C_1\right]^2 = \frac{3}{16}(a+b+c)(-a+b+3c)(-b+c+3a)(-c+a+3b)$$
Since $a, b, c$ are the sides of a triangle, there are positive real numbers $p, q, r$ such that $a = q + r$, $b = r + p$, $c = p + q$. Using these relations we obtain
$$\frac{[ABC]^2}{\left[A_1 B_1 C_1\right]^2} = \frac{16pqr}{3(2p+q)(2q+r)(2r+p)}$$
Thus it is sufficient to prove that
$$(2p+q)(2q+r)(2r+p) \geq 27pqr$$
for positive real numbers $p, q, r$. Using AM-GM inequality, we get
$$2p+q \geq 3(p^2 q)^{1/3}, \quad 2q+r \geq 3(q^2 r)^{1/3}, \quad 2r+p \geq 3(r^2 p)^{1/3}$$
Multiplying these, we obtain the desired result. We also observe that equality holds if and only if $p = q = r$. This is equivalent to the statement that $ABC$ is equilateral.