Nested radical of Ramanujan and cyclic inequality Inspired by a nested radical of Ramanujan I propose this :

Let $a,b,c\in (2\sqrt[4\,]{5},6-2\sqrt[4\,]{5})$ such that $a+b+c=9$ then we have :
  $$\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{c+2\sqrt[4\,]{5}}{c-2\sqrt[4\,]{5}}}\leq\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{c-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{c+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}$$

I try to prove it term by term like this :
$$\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}\leq \sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}$$
But it's cyclic so we are in the wrong way.
On the other hand I can prove that :
$$3\sqrt[4\,]{\frac{3+2\sqrt[4\,]{5}}{3-2\sqrt[4\,]{5}}}\leq\sqrt[4\,]{\frac{a+2\sqrt[4\,]{5}}{a-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{b+2\sqrt[4\,]{5}}{b-2\sqrt[4\,]{5}}}+\sqrt[4\,]{\frac{c+2\sqrt[4\,]{5}}{c-2\sqrt[4\,]{5}}}$$
Because it's not hard to show that the function $f(x)=\sqrt[4\,]{\frac{x+2\sqrt[4\,]{5}}{x-2\sqrt[4\,]{5}}}$ is convex on the interval $(2\sqrt[4\,]{5},6-2\sqrt[4\,]{5})$ and after apply Jensen's inequality.
I tried also majorization ,C-S,AM-GM and more but without success.
If you have nice idea I let you play with this .
Thanks a lot for sharing your time and knowledge .
 A: Remark: For applications of the Symmetric Function Theorem, also see
Prove that $\sum\limits_{cyc}\frac{1}{\sqrt{a^2+ab+b^2}}\geq\frac{2}{\sqrt{ab+ac+bc}}+\sqrt{\frac{a+b+c}{3(a^3+b^3+c^3)}}$
How prove this inequality $\sum\limits_{cyc}\frac{x+y}{\sqrt{x^2+xy+y^2+yz}}\ge 2+\sqrt{\frac{xy+yz+xz}{x^2+y^2+z^2}}$
Proof: We will apply Ji Chen's Symmetric Function Theorem for $n=3$
(see https://artofproblemsolving.com/community/c6h194103p1065812):
Symmetric Function Theorem: Let $d\in (0,1)$. Let $x, y, z, u, v, w$ be non-negative real numbers satisfying
$$x+y+z \ge u+v+w, \quad
xy+yz+zx \ge uv+vw+wu, \quad
xyz \ge uvw.$$
Then $x^d + y^d+z^d \ge u^d + v^d+w^d$.
Now let us prove the inequality. We need to prove that
$$\sqrt[4]{X} + \sqrt[4]{Y} + \sqrt[4]{Z} \le \sqrt[4]{U} + \sqrt[4]{V} + \sqrt[4]{W}.$$
Clearly, $XYZ = UVW$. According to the Symmetric Function Theorem, it suffices to prove that 
$$X + Y + Z \le U + V + W$$
and
$$XY + YZ + ZX \le UV + VW + WU.$$
After some manipulations, it suffices to prove that
$$2(-a^2+ab+ac-b^2+bc-c^2)\sqrt[4]{5} + a^2c+ab^2-3abc+bc^2\ge 0$$
and
$$2(a^2-ab-ac+b^2-bc+c^2)\sqrt[4]{5} +a^2b-3abc+ac^2+b^2c \ge 0.$$
The latter is obvious.
I used Mathematica Resolve to verify the former. I believe that the proof of the former is not very hard (omitted here).
We are done.
