Prove $\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1$ and $ \sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1$ for $x,y,z>0,xyz=1$ Source: https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3
Let $x,y,z>0,xyz=1$ then $$
\sum_{cyc}{\sqrt{\frac{x+1}{x^2+16x+1}}}\geqslant 1  \ \ \ \ (1)
$$
and$$
\sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\leqslant 1\ \ \ (2)
$$
For this type of inequalities, I usually use local inequalities$$
x,y,z>0,xyz=1,\sum_{cyc}{\frac{1}{x^2+x+1}}\geqslant 1, \sum_{cyc}{\frac{x^2}{x^2+x+1}}\geqslant 1, \sum_{cyc}{\frac{x+1}{x^2+x+1}}\leqslant 2
$$
For (1) it's smooth,because$$
\frac{x^4+1}{x^8+16 x^4+1}-\left(\frac{1}{x^2+x+1}\right)^2=\frac{(x-1)^2 x \left(2 x^4+7 x^3+14 x^2+7 x+2\right)}{\left(x^2+x+1\right)^2 \left(x^8+16 x^4+1\right)} \geqslant 0$$
Then it's easy to prove.
But for (2) I get
$$\frac{x^2+1}{4 x^4+10 x^2+4}-\left(\frac{x+1}{2 \left(x^2+x+1\right)}\right)^2=\frac{(x-1)^2 x^2}{4 \left(x^2+2\right) \left(x^2+x+1\right)^2 \left(2 x^2+1\right)} \geqslant 0$$
Then $$
\sum_{cyc}{\sqrt{\frac{x+1}{4x^2+10x+4}}}\geqslant \sum_{cyc}{\frac{1}{2}\frac{\sqrt{x}+1}{x+\sqrt{x}+1}}\leqslant 1
$$
Failed. How can I prove (2)? Any solution is welcome.
 A: Hint :
For $x\leq 0$ we have :
$$\frac{d}{dx}\left(\frac{d}{dx}f\left(e^{x}\right)\right)<0$$
Where :
$$f(x)=\sqrt{\frac{x+1}{4x^{2}+4+10x}}$$
Next we show that for $x>0$ and $n= 3$ a natural number :
$$f\left(x\right)+\left(n-1\right)f\left(\frac{1}{x^{\frac{1}{n-1}}}\right)-nf\left(1\right)\leq 0$$
In the case $n=3$ we can use the bound given in the link https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3
We have for $x>0$:
$$\frac{1}{1+\sqrt{1+3x}}-\sqrt{\frac{x+1}{4x^{2}+4+10x}}\geq 0$$
To show this inequality we use the substitution $y^2=3x+1$ then square both side then clearing the denominator and use a factorization .
So we need to show :
$$\frac{1}{1+\sqrt{1+3x}}+2\frac{1}{1+\sqrt{1+3x^{-0.5}}}-3\frac{1}{1+\sqrt{1+3}}\leq 0$$
Wich is easy !
So now we can apply the LCF corollary (first reference) and conclude .
Last remark : We have also the  case $n=4$ for $a_i\in(e^{-1},e^{1})$ such that $\prod_{i=1}^{4}a_i=1$.
Reference :

*

*Cirtoaje and Baiesu: An extension of Jensen’s discrete inequality to half convex functions.
Journal of Inequalities and Applications 2011 2011:101.

2)https://artofproblemsolving.com/community/c6t243f6h2745656_inequalities_with_3



Edit 11/02/2022 :
For $0<x\leq 1$ we have :
$$\frac{\left(x+2\right)\left(x+1\right)}{\frac{11}{2}x^{2}+4+\frac{17}{2}x}\geq f(x)$$
On the other hand we have $x\geq 1$ :
$$h\left(x\right)=\frac{2}{3}-\frac{1}{1.5x^{-0.5}+1.5}\geq f(x)$$
Edit 12/02/2022 :
We can do better on $x\in (0,1]$ we have :
$$r(x)=\frac{\frac{1}{2}x^{2}+\frac{7}{2}x+2}{4x^{2}+4+10x}\geq f(x)$$
Using the same approach as above we have on $(-\infty,0]$ :
$$\frac{d}{dx}\left(\frac{d}{dx}r\left(e^{x}\right)\right)<0$$
So we have for $0<a\leq 1$ and $0<b\leq 1$ and $c\geq 1$ :
$$f(a)+f(b)+f(c)\leq 2r(\sqrt{ab})+h(c)=2r(\sqrt{ab})+h(1/ab)\leq 1$$
Wich is smooth .
For the other case I use Buffalo's ways see Wolfram alpha
*I made a mistake I solve another problem ...
Here you can find the solution for the other case wich use Buffalo's way
See Wolfram alpha (2).
