Proof.
The desired inequality is written as
$$\sqrt{\frac{2a + 3bc}{5}} + \sqrt{\frac{2b + 3ca}{5}} + \sqrt{\frac{2c + 3ab}{5}} \ge 3. \tag{1}$$
By AM-GM, we have
$$\sqrt{\frac{2a + 3bc}{5}} = \frac{12(2a + 3bc)}{2\sqrt{180(2a + 3bc)}}
\ge \frac{12(2a + 3bc)}{\frac{180(2a + 3bc)}{25 + 5bc} + (25 + 5bc)}. \tag{2}$$
From (1) and (2), it suffices to prove that
$$\sum_{\mathrm{cyc}} \frac{12(2a + 3bc)}{\frac{180(2a + 3bc)}{25 + 5bc} + (25 + 5bc)} \ge 3. \tag{3}$$
We use the pqr method.
Let $p = a + b + c, q = ab + bc + ca = 3, r = abc$. We have
$$p \ge 3, \quad 0 \le r \le 1, \quad r \ge \frac{12p - p^3}{9}. \tag{4}$$
(Note: $r \ge \frac{12p - p^3}{9}$ follows from the degree three Schur inequality $r \ge \frac{4pq - p^3}{9}$.)
(3) is equivalently written as
$$775r^4 + m_3 r^3 + m_2 r^2 + m_1 r + m_0 \ge 0 \tag{5}$$
where
\begin{align*}
m_3 &= 200p^2 + 16510p + 3360, \\
m_2 &= 7840p^3 + 6115p^2 - 4720p + 672952, \\
m_1 &= 788768p^2 - 1388944p - 3990360, \\
m_0 &= 16000p + 29308.
\end{align*}
It suffices to prove that (5) is true for all $p, r$ satisfying (4).
Since $m_3, m_2, m_0 \ge 0$, we only need to prove the case that $m_1 < 0$ which implies $p < 2\sqrt{3}$. Thus, $r \ge \frac{12p - p^3}{9} \ge 0$. Let $r_1 := \frac{12p - p^3}{9}$. We have
\begin{align*}
&775r^4 + m_3 r^3 + m_2 r^2 + m_1 r + m_0\\
\ge{}& 775r_1^4 + m_3 r_1^3 + m_2 r_1^2 + m_1 + m_0 \\
={}& \frac{1}{6561}(p - 3)g(p)\\
\ge{}& 0
\end{align*}
where
\begin{align*}
g(p) &:= 775\,{p}^{11}+525\,{p}^{10}-184215\,{p}^{9}+116955\,{p}^{8}+6865020\,{
p}^{7}\\
&\qquad +5282820\,{p}^{6}-11077668\,{p}^{5}+57435156\,{p}^{4}-791753940
\,{p}^{3}\\
&\qquad -2378061180\,{p}^{2}+5890235436\,p+8662820724.
\end{align*}
We are done.
