Cyclic inequality. $\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$ Let $x,y,z>0$. Show that
$$\sum_{cyc} \frac{x+2y}{\sqrt{z(x+2y+3z)}}\geq \frac{3\sqrt{6}}{2}$$
Equality case is for $x=y=z$.
A hint I have is that I have to amplify with something convenient, and then apply some sort of mean inequality inequality. Also, how would one go about finding with what to amplify?
I've tried amplifying the given fraction with a lot of random stuff, but no succes in finding something where I can apply the AM $>$ GM inequality.
 A: By AM-GM and Cauchy-Schwarz we obtain:
$$\sum_{cyc}\frac{x+2y}{\sqrt{z(x+2y+3z}}=\sum_{cyc}\frac{2\sqrt{6}(x+2y)}{2\sqrt{6z(x+2y+3z})}\geq2\sqrt6\sum_{cyc}\frac{x+2y}{x+2y+9z}=$$
$$=2\sqrt6\sum_{cyc}\frac{(x+2y)^2}{(x+2y)^2+9(x+2y)z}\geq2\sqrt6\cdot\frac{\left(\sum\limits_{cyc}(x+2y)\right)^2}{\sum\limits_{cyc}((x+2y)^2+9z(x+2y))}=$$
$$=\frac{18\sqrt6(x+y+z)^2}{\sum\limits_{cyc}(5x^2+31xy)}\geq\frac{3\sqrt6}{2},$$
where the last inequality it's just $$\sum_{cyc}(x-y)^2\geq0.$$
A: Another way.
By Holder $$\sum_{cyc}\frac{x+2y}{\sqrt{z(x+2y+3z)}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x+2y}{\sqrt{z(x+2y+3z)}}\right)^2\sum\limits_{cyc}(x+2y)z(x+2y+3z)}{\sum\limits_{cyc}(x+2y)z(x+2y+3z)}}\geq$$
$$\geq\sqrt{\frac{27(x+y+z)^3}{\sum\limits_{cyc}(7x^2y+7x^2z+4xyz)}}\geq\frac{3\sqrt6}{2},$$
where the last inequality it's $$\sum_{cyc}(2x^3-x^2y-x^2z)\geq0$$ or
$$\sum_{cyc}(x-y)^2(x+y)\geq0.$$
A: Apply AM-GM inquality now and Cauchy-Schwarz inequalities ( at later step ) we have: $\displaystyle \sum_{\text{cyclic}} \dfrac{x+2y}{\sqrt{6z(x+2y+3z)}}\ge \displaystyle \sum_{\text{cyclic}} \dfrac{2(x+2y)}{6z+(x+2y+3z)}= \displaystyle \sum_{\text{cyclic}} \dfrac{2(x+2y)}{(x+2y)+9z}$. Next, let $a = x+2y, b = y + 2z, c = z + 2x$, then solve for $x,y,z$ in terms of $a,b,c$ we have: and the above cyclic sum becomes: $\displaystyle \sum_{\text{cyclic}} \dfrac{2a}{4b-a+c} = \displaystyle \sum_{\text{cyclic}} \dfrac{2a^2}{4ab-a^2+ac} \ge \dfrac{2(a+b+c)^2}{5ab+5bc+5ca - a^2-b^2-c^2}\ge \dfrac{3}{2}\iff a^2+b^2+c^2 \ge ab+bc+ca \iff (a-b)^2+(b-c)^2+(c-a)^2 \ge 0$, which is clearly true. So the inequality is proven and $=$ occurs when $a = b = c $ or $x = y = z$.
Note: Observe that the denominators of the cyclic sum above are positive each because $4b - a + c = (x+2y) + 9z > 0$ since $x,y,z > 0$. And also: $x = \dfrac{a+4c-2b}{9}, y = \dfrac{b+4a-2c}{9}, z = \dfrac{c+4b-2a}{9}$.
