Let $x$, $y$, $z$ be positive real numbers such that $a = x/y$, $b = y/z$, $c = z/x$. After substitution and simplifying the given inequality is transformed to the following one:
$$4(x + y + z)^{\sqrt{3}xyz} \le 3(x + y)^{2/3}(y + z)^{2/3}(z + x)^{2/3}$$
For triangle with the sides lengths $u = x + y$, $v = y + z$, $w = z + x$ the last inequality gives
$$4p\sqrt[3]{(p-u)(p-v)(p-w)} \le 3(uvw)^{2/3}$$
where $p$ is a semiperimeter of the triangle. By Heron's formula we obtain that the inequality is equivalent to:
$$(2p)^{2/3} \le (3\sqrt{3}\frac{uvw}{4s})^{2/3} = (3\sqrt{3}R)^{2/3}$$
where $S$ is area and $R$ is a radius of the circumcircle of the triangle. But the inequality
$$u + v + w = 2p \le 3\sqrt{3}R$$
is well-known; there are several ways to prove that.

By raising to the third power the given inequality and taking into account the condition $abc = 1$ we eliminate radicals. After simplifying we get the following inequality:
$$114 + 30 \sum_{cyc} \left(a + \frac{1}{a}\right) + 10 \sum_{cyc} \frac{a}{b} \le 54 \sum_{cyc} \frac{b}{a} + 27 \sum_{cyc} \left(a^2 + \frac{1}{a^2}\right)$$
The last inequality can be obtained from the following elementary inequalities:
$$2 \sum_{cyc} \frac{a}{b} \le \sum_{cyc} \left(a^2 + \frac{1}{a^2}\right), \quad 3 \le \sum_{cyc} \frac{b}{a},$$
$$a + \frac{1}{a} \le \frac{1}{4} \left(a^2 + \frac{1}{a^2} + 6\right), \quad 2 \le a^2 + \frac{1}{a^2}.$$