Prove: $\sqrt{\frac{bc}{a(3b+a)}} + \sqrt{\frac{ac}{b(3c+b)}} + \sqrt{\frac{ab}{c(3a+c)}} \ge \frac{3}{2}$. Prove: $\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{3}{2}$ with $a, b, c$ are positive real numbers.
Let $a \le b \le c$:
\begin{align*}
\sqrt{\dfrac{bc}{a(3b+a)}} = \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{c}{3b+a}} &\ge \sqrt{\dfrac{b}{a}}\sqrt{\dfrac{c}{3c+c}} = \dfrac{1}{2} \sqrt{\dfrac{b}{a}} && (1)
\\
\sqrt{\dfrac{ac}{b(3c+b)}} = \sqrt{\dfrac{a}{b}}\sqrt{\dfrac{c}{3c+b}} &\ge \sqrt{\dfrac{a}{b}}\sqrt{\dfrac{c}{3c+c}} = \dfrac{1}{2}\sqrt{\dfrac{a}{b}} && (2)
\\
\sqrt{\dfrac{ab}{c(3a+c)}} \ge \sqrt{\dfrac{ab}{c(3c+c)}} &\ge \dfrac{1}{2} \dfrac{\sqrt{ab}}{c} && (3)
\end{align*}
With $(1)+(2)+(3)$, we have:
$$
\sqrt{\dfrac{bc}{a(3b+a)}} + \sqrt{\dfrac{ac}{b(3c+b)}} + \sqrt{\dfrac{ab}{c(3a+c)}} \ge \dfrac{1}{2}\left(\sqrt{\dfrac{b}{a}} + \sqrt{\dfrac{a}{b}} + \dfrac{\sqrt{ab}}{c} \right).
$$
And I can not  find the method to finish this problem. I am trying to think another method.
 A: From an old Vasile Cirtoaje inequality (see solution here, or here)

If $x,\,y,\,z$ are positive numbers, such that $xyz=1,$ then $$\sqrt{\frac{x}{y+3}}+\sqrt{\frac{y}{z+3}}+\sqrt{\frac{z}{x+3}}\geqslant\frac{3}{2}. \quad (1)$$

Beause $abc=1,$ we can put $x=\frac ab,\,y = \frac bc, \, z=\frac ca.$ Now, the inequality $(1)$ become
$$\sqrt{\frac{bc}{a(a+3b)}} + \sqrt{\frac{ac}{b(b+3c)}} + \sqrt{\frac{ab}{c(c+3a)}} \geqslant \frac{3}{2}.$$
A: After replacing $a$ at $\frac{1}{a},$ $b$ at $\frac{1}{b}$ and $c$ at $\frac{1}{c}$ we need to prove that:
$$\sum_{cyc}\frac{a}{\sqrt{c(3a+b)}}\geq\frac{3}{2}.$$
Now, by Holder $$\left(\sum_{cyc}\frac{a}{\sqrt{c(3a+b)}}\right)^2\sum_{cyc}ac(3a+b)\geq(a+b+c)^3.$$
Id est, it's enough to prove that
$$4(a+b+c)^3\geq9\sum_{cyc}(3a^2c+abc),$$ which is smooth:
Let $\{a,b,c\}=\{x,y,z\}$ where $x\geq y\geq z$.
Thus, by Rearrangement and AM-GM we obtain:
$$9\sum_{cyc}(3a^2c+abc)=27(a\cdot ac+b\cdot ba+c\cdot cb+xyz)\leq$$
$$\leq27(x\cdot xy+y\cdot xz+z\cdot yz+xyz)=27y(x+z)^2=108y\left(\frac{x+z}{2}\right)^2\leq$$
$$\leq108\left(\frac{y+\frac{x+z}{2}+\frac{x+z}{2}}{3}\right)^3=4(x+y+z)^3=4(a+b+c)^3.$$
