$\sum\limits_{cyc} \sqrt{\frac{a^3}{1+bc}} \geq 2$ for $a, b, c > 0$ which satisfies $abc=1$. $\displaystyle \sum_{cyc} \sqrt{\dfrac{a^3}{1+bc}} \geq 2$ for $a, b, c > 0$ which satisfies $abc=1$.
My attempt:
\begin{align}
&\text{let } a=\frac{y}{x}, b=\frac{x}{z}, c=\frac{z}{y}. \\
&\text{Substituting for the original F.E.: }\displaystyle \sum_{cyc}\sqrt{\frac{(\frac{y}{x})^3}{1+\frac{x}{y}}} \geq 2 \text{ for }x, y, z\in \mathbb{R}^+. \\
&\therefore \text{ETS) }\displaystyle \sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}} \geq 2. \\
\ \\
& \text{Two ways to think: }\\
\ \\
& (1) \\
&\therefore \text{Using Cauchy-Schwarz inequality, } \displaystyle \Bigg(\sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}}\Bigg) \Bigg(\sum_{cyc} \frac {x}{y}\Bigg) \geq \sum_{cyc} \frac{y^2}{x^2+xy} \\
&\text{ETS) }\displaystyle \sum_{cyc} \frac{y^2}{x^2+xy} \geq 2\Bigg( \sum_{cyc} \frac {x}{y} \Bigg).
\ \\
&(2)\\
&\therefore \text{Using Cauchy-Schwarz inequality, } \displaystyle \Bigg(\sum_{cyc} \sqrt{\frac{y^4}{x^3y+x^4}}\Bigg) \Bigg(\sum_{cyc} \sqrt{\frac {x+y}{y}}\Bigg) \geq \sum_{cyc} \frac{y^3}{x^3} \\
&\text{ETS) }\displaystyle \sum_{cyc} \frac{y^3}{x^3} \geq 2\Bigg( \sum_{cyc} \sqrt{\frac{x+y}{y}} \Bigg).
\end{align}
p.s. I think we can't use the AM-GM one, but I'll try.
 A: Using Titu's Lemma,
$$\sum\sqrt{\frac{a^3}{1+bc}}=\sum \frac{a^2}{\sqrt{a+1}}\ge\frac{(a+b+c)^2}{\sum \sqrt{a+1}} \tag{1}$$
It remains to show,
$$(a+b+c)^2\ge 2(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1})$$
Using A.M-G.M inequality,
$$\sum \frac{(a+1)+1}{2}\ge \sum\sqrt{(a+1)\cdot 1}\implies a+b+c+6\ge2(\sqrt{a+1}+\sqrt{b+1}+\sqrt{c+1}) $$
It remains to show,
$$(a+b+c)^2 \ge (a+b+c)+6\tag{2}$$
which is true, since $a+b+c\ge 3\sqrt[3]{abc}=3.$

 Proof of $(2)$:$$(a+b+c)^2\ge 3(a+b+c)\ge (a+b+c)+6$$


Given below is a proof of a stronger result.
Claim:
$$\sum \sqrt{\frac{a^3}{1+bc}}\ge \frac{3}{\sqrt{2}}$$
Proceeding from $(1)$, it remains to show,
$$\sqrt{2}(a+b+c)^2\ge3\sum{\sqrt{a+1}}\implies 4(a+b+c)^2\ge 3\sum2\sqrt{2(a+1)}$$
Using A.M-G.M inequality,
$$\frac{\sum(a+1)+2}{2}\ge \sum \sqrt{2(a+1)}\implies a+b+c+9\ge \sum2\sqrt{2(a+1)} $$
It remains to show,
$$4(a+b+c)^2\ge 3(a+b+c)+27$$
which is true, since $a+b+c\ge 3.$
Also, equality can be achieved, unlike the original problem, at $a=b=c=1$.
