If $abc=1$ so $\sum\limits_{cyc}\sqrt{\frac{a}{4a+2b+3}}\leq1$. Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\sqrt{\frac{a}{4a+2b+3}}+\sqrt{\frac{b}{4b+2c+3}}+\sqrt{\frac{c}{4c+2a+3}}\leq1$$
The equality "occurs" also for $a\rightarrow+\infty$, $b\rightarrow+\infty$ and $a>>b$.
I tried AM-GM, C-S and more, but without any success. 
 A: We begin with a theorem :
Theorem :
Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $d\geq e \geq f $ under the three following conditions :
$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :
$$a+b+c\geq d+e+f$$
Here we suppose that we have :
$a\geq b \geq 1 \geq c $
So to get the majorization we prove this :
$\sqrt{\frac{a}{4a+2b+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$
Wich is equivalent to :
$\frac{a}{4a+2b+3}\geq \frac{c}{4c+2a+3}$
Or :
$a(4c+2a+3)\geq c(4a+2b+3)$
Wich is obvious under the previous conditions.
With the same reasoning we can prove that we have :
$\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$
Now we study the case :
$\sqrt{\frac{a}{4a+2b+3}}\geq\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem 
And
$0.5-\frac{1}{8.2a}\geq 0.5-\frac{1}{8.2b}\geq \frac{1}{8.2a}+\frac{1}{8.2b} $
wich corresponding to $d\geq e \geq f $ in the initial theorem 
It's clear that we have :
$\sqrt{\frac{a}{4a+2b+3}}\leq \sqrt{\frac{a}{4a+3}}\leq 0.5-\frac{1}{8.2a}$
And 
$\sqrt{\frac{b}{4b+2c+3}}\leq \sqrt{\frac{b}{4b+3}}\leq 0.5-\frac{1}{8.2b}$
So we have :
$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})$
And 
$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\sqrt{\frac{c}{4c+2a+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})(\frac{1}{8.2a}+\frac{1}{8.2b})$
Wich is true because we have with the condition $abc=1$
$$27\leq \prod_{cyc}\sqrt{4a+2b+3}$$
So now you just have to apply the theorem with this majorization .
The case $\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{a}{4a+2b+3}} \geq \sqrt{\frac{c}{4c+2a+3}}$ is the same.
And for the case $a\geq 1 \geq b \geq c$ you just have to make the following substitution $B=\frac{1}{b}$ to find the previous case $a\geq b \geq 1 \geq c $
Edit :
With the previous substitution the original inequality becomes with $a\geq b \geq 1 \geq c$ and $ac=b$:
$$\sqrt{\frac{ab}{4ab+2+3b}}+\sqrt{\frac{1}{4+2cb+3b}}+\sqrt{\frac{c}{4c+2a+3}}$$
We can briefly prove that we have :
$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}$
And 
$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$
Now we study the case :
$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem 
And 
$0.5-\frac{1}{11(ab)^2}\geq \frac{1}{3} \geq 1-\frac{1}{3}-(0.5-\frac{1}{11(ab)^2})$
wich corresponding to $d\geq e \geq f $ in the initial theorem 
Now you just have to apply the theorem with this majorization .
The case $\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}\geq \sqrt{\frac{1}{4+2cb+3b}}$ works this the same majorization. 
