Proof.
The proof was inspired by Michael Rozenberg's idea.
Firstly, we use Cauchy-Schwarz as \begin{align*} (a+b+c+1)\sqrt{a+3}&=\sqrt{[a(1+b+c)+bc][4a(1+b+c)+(b+c+1-a)^2]}\\&\ge 2a(1+b+c)+\sqrt{bc}(b+c+1-a), \end{align*} which implies $$(a+b+c+1)\left(\sqrt{a+3}+\sqrt{bc}\right)\ge 2(b+c+1)(a+\sqrt{bc}),$$or$$\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge \frac{b+c+1-a}{a+b+c+1}+\frac{a}{a+\sqrt{bc}}.\tag{*}$$ Now, sum cycliclycyclically on $(*)$ we obtain $$\sum_{cyc}\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{a+b+c+3}{a+b+c+1}+\sum_{cyc}\frac{a}{a+\sqrt{bc}}. $$ The last inequality is true by Cauchy-Schwarz and AM-GM and the desired result follows.
Indeed, by using Cauchy-Schwarz inequality$$\sum_{cyc}\frac{a}{a+\sqrt{bc}}\ge \frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}.$$ Also, it is easily $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le 3.$ Hence, apply AM-GM $$\sum_{cyc}\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{a+b+c+3}{a+b+c+1}+\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2}{a+b+c+3}\ge\frac{2\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}{\sqrt{a+b+c+1}} . $$ The proof is done. Equality holds at $a=b=c=1.$
Motivation.
The key step is $(*).$
When I tried to multiply $a+b+c+1$ to $LHS$ of the OP, I came up with Cauchy-Schwarz 's idea.