Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels like nothing simple fits. Next I noticed we can normalise: restricting $a+b+c=1$ it can be made to look like this:
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
Another idea is to let $x=a/b,y=b/c, z=c/a:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{1}{x(y+1)}}+\sqrt{\frac{1}{y(z+1)}}+\sqrt{\frac{1}{z(x+1)}}\right)\leq \sqrt{x+y+z}$$
But I can't see where to go from here.

After using C-S $\left(\sum\limits_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq(a+b+c)\sum\limits_{cyc}\frac{1}{b+c}$ we'll obtain something obvious:
$$\sum_{cyc}(3a^4c^2+3a^3b^3+a^4bc-2a^3b^2c-2a^3c^2b-3a^2b^2c^2)\geq0.$$
Done!
