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.
 A: Using C-S inequality, we can obtain 
$$\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)^2\leq3\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).$$
Since $(b+c)\left(\frac1{b}+\frac1{c}\right)\geq4$, we have $\frac{a}{b+c}\leq\frac1{4}\left(\frac{a}{b}+\frac{a}{c}\right)$, similarly we can have
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{4}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right).$$
So $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}$ must be less than or equal to either $\frac1{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)$ or $\frac1{2}\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right)$. If $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{2}\left(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\right),$$
then we may exchange $b$ and $c$, hence we can always get the inequality $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\leq\frac1{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right).$$ Therefore 
$$\sqrt{\frac2{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq\sqrt{2}\sqrt{\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}}\\\leq\sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}.$$   
A: I can obtain something a bit weaker. Since the square root is a concave fucntion, we have that
$$
1/3(\sqrt{A}+\sqrt{B}+\sqrt{C})\leq \sqrt{1/3 (A+B+C)}.
$$
This implies
$$
\frac{\sqrt{3}}{3}\left(\sqrt{A}+\sqrt{B}+\sqrt{C}\right)\leq \sqrt{A+B+C}.
$$
Now, take $A=a/(b+c)$, $B=b/(a+c)$ and $C=c/(b+a)$. Since $a,b,c>0$ we have $A<a/b$, $B<b/c$ and $C<c/a$. Then
$$
\frac{1}{\sqrt{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}.
$$
A: 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!
