Rewrite the left hand side of inequality in following way:
$$\frac{a+b+c+3}{4} = \frac{a+b+c+3}{4\sqrt{abc}} = \frac{a+1}{4\sqrt{abc}} + \frac{b+1}{4\sqrt{abc}} + \frac{c+1}{4\sqrt{abc}}$$
Rewrite denominators:
$$\frac{a+1}{4\sqrt{abc}} + \frac{b+1}{4\sqrt{abc}} + \frac{c+1}{4\sqrt{abc}} = \frac{a+1}{2\sqrt{ab \cdot c} + 2\sqrt{ac \cdot b}} + \frac{b+1}{2\sqrt{ba \cdot c} + 2\sqrt{bc \cdot a}} + \frac{c+1}{2\sqrt{ca \cdot b} + 2\sqrt{cb \cdot a}} =$$
and then by arithmetic mean-geometric mean inequality, we have
$$\begin{aligned} &= \frac{a+1}{ab+c+ac+b} + \frac{b+1}{bc+a+ba+c} + \frac{c+1}{ca+b+cb+a} = \frac{a+1}{(a+1)(b+c)} + \frac{b+1}{(b+1)(a+c)} + \frac{c+1}{(c+1)(b+a)} = \\&= \frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{b+a} = \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}. \end{aligned}$$

We introduce change of variables: $x=a^3, y=b^3, z=c^3$. We now have the condition $xyz=1$.
We apply Schur inequality (with exponent $r=1$) to the numerator of the left hand side:
$$x^3 + y^3 + z^3 + 3xyz \ge x^2y + x^2z + y^2x + y^2z + z^2x + z^2y$$
to obtain inequality
$$\frac{x^2y + x^2z + y^2x + y^2z + z^2x + z^2y}{4} \ge \frac{1}{x^3+y^3} + \frac{1}{y^3+z^3} + \frac{1}{z^3+x^3}.$$
We apply arithmetic mean-geometric mean inequality for the denominators of the right hand side:
$$x^3 + y^3 \ge 2x^{3/2}y^{3/2} \quad \Rightarrow \quad \frac{1}{x^3 + y^3} \le \frac{1}{2x^{3/2}y^{3/2}} = \frac{1}{2}z^2\sqrt{xy}$$
and similarly to the other terms. We now have to prove
$$\frac{x^2y + x^2z + y^2x + y^2z + z^2x + z^2y}{4} \ge \frac{1}{2}x^2\sqrt{yz} + \frac{1}{2}y^2\sqrt{xz} + \frac{1}{2}z^2\sqrt{xy}$$
$$\frac{x^2y + x^2z + y^2x + y^2z + z^2x + z^2y}{2} \ge x^2\sqrt{yz} + y^2\sqrt{xz} + z^2\sqrt{xy}.$$
We apply arithmetic mean-geometric mean inequality in pairs on the left hand side:
$$\frac{x^2y + x^2z}{2} \ge x^2\sqrt{yz}$$
$$\frac{y^2z + y^2x}{2} \ge y^2\sqrt{xz}$$
$$\frac{z^2x + z^2y}{2} \ge z^2\sqrt{xy}.$$
Summing up inequalities from above finishes the proof.