Solution 1. By Holder Inequality we have
$$\left(\sum_{cyc} \frac{a\sqrt{a}}{bc}\right)\left(\sum_{cyc} bc\right)\left(\sum_{cyc} 1\right) \geq \left(\sum_{cyc} \sqrt{a}\right)^3 \Rightarrow \sum_{cyc} \frac{a\sqrt{a}}{bc} \geq \frac{1}{3}\left(\sum_{cyc} \sqrt{a}\right)^3.$$
So it suffices to have
$$\frac{1}{3}\left(\sum_{cyc} \sqrt{a}\right)^3 \geq \sqrt{3}\left(\sum_{cyc} \sqrt{a}\right) \Leftrightarrow \left(\sum_{cyc} \sqrt{a}\right)^2 \geq 3\sqrt{3} \Leftrightarrow \sum_{cyc} \sqrt{a} \geq \sqrt[4]{27}$$
Now suppose that $\sum_{cyc} \sqrt{a} < \sqrt[4]{27}$. It suffices to prove that $\sum_{cyc} \frac{a\sqrt{a}}{bc} \geq \sqrt{3\sqrt[4]{27}}$, but
we have $\sum_{cyc} \frac{a\sqrt{a}}{bc} \geq 3\sqrt[3]{\prod_{cyc} \frac{a\sqrt{a}}{bc}} = 3(abc)^{-\frac{1}{6}}$ (*) by AM-GM inequality
$$\begin{align*} 1 &= \sum_{cyc} ab \geq 3(abc)^{\frac{2}{3}} \Rightarrow (abc)^{-\frac{2}{3}} \geq 3 \Rightarrow (abc)^{-\frac{1}{6}} \geq \sqrt[4]{3} \\ \Rightarrow \sum_{cyc} \frac{a\sqrt{a}}{bc} &\geq 3(abc)^{-\frac{1}{6}} \geq 3\sqrt[4]{3} = 3^{\frac{5}{4}} = 3^{\frac{1}{2}} \cdot 3^{\frac{3}{4}} = \sqrt{3\sqrt[4]{27}}. \end{align*}$$

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Solution 2. First by Cauchy-Schwarz inequality we have
$$\left(\sum_{cyc} \frac{a^2}{bc\sqrt{a}}\right)\left(\sum_{cyc} bc\sqrt{a}\right) \geq \left(\sum_{cyc} a\right)^2.$$
So it suffices to prove that
$$\left(\sum_{cyc} a\right)^2 \sqrt{\sum_{cyc} ab} \geq \sqrt{3} \times \sqrt{abc} \left(\sum_{cyc} \sqrt{bc}\right)\left(\sum_{cyc} \sqrt{a}\right).$$
Since $\sum_{cyc} ab = 1$.
But by AM-GM and Cauchy-Schwarz we have
$$\sqrt{\sum_{cyc} ab} \geq \frac{1}{\sqrt{3}} \sum_{cyc} \sqrt{ab} \quad (\text{Cauchy-Schwarz}) \quad (1)$$
$$\sum_{cyc} a \geq \frac{1}{3} \left( \sum_{cyc} \sqrt{a} \right)^2 \quad (\text{Cauchy-Schwarz}) \quad (2)$$