Hi it's a follow up of 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 $x,y,z>0,xyz=1$ :
Problem :
Let $x_i>0$ such that $\prod_{i=1}^{n}x_i=1$ and $n\geq 3$
Then prove or disprove that :
$$\sum_{i=1}^{n}\sqrt{\frac{x_i+1}{4x_i^2+10x_i+4}}\leq \frac{n}{3}$$
My attempt :
In the giving link I give a skectch of proof for the case $n=3$ using the simple (I mean easy) bounds :
$x\in (0,1]$ we have :
$$r(x)=\frac{\frac{1}{2}x^{2}+\frac{7}{2}x+2}{4x^{2}+4+10x}\geq \sqrt{\frac{x+1}{4x^2+10x+4}}$$
And for $x\geq 1$ :
$$h(x)=\frac{2}{3}-\frac{1}{1.5x^{-0.5}+1.5}\geq\sqrt{\frac{x+1}{4x^2+10x+4}}$$
Other case :
For the general case and using the fact :
Let $x\in (-\infty,0]$ then we have :
$$\frac{d}{dx}\left(\frac{d}{dx}r\left(e^{x}\right)\right)<0$$
So we can use Jensen's inequality for $0<x_k\leq 1$ so we need to show :
$$\sum_{i=1}^{n-k}h(x_i)+k r\left(\prod_{i=n-k+1}^{n}(x_i)^{\frac{1}{k}}\right)\leq \frac{n}{3}$$
If there is some mistake related to the index tell me .
For example in the case $n=4$ when two variables are superior to one and the two other less than one I use Buffalo's way see Wolfram alpha .
Edit :
In fact in the WA link I show something stronger .
Questions :
1)How to (dis)prove it ?
2)How to use Buffalo's way in the general case (if true)?
Thanks in advance .
$x=0.1$ gives $$\sqrt\frac{1.1}{0.04+1+4} > 0.46$$ Therefore we can have $x_i = 0.1$ for $i \in [1;n-1]$ and $$\sum_{i=1}^n\sqrt\frac{x_i+1}{4x_i^2+10x_i+4} > \sum_{i=1}^{n-1}\sqrt\frac{x_i+1}{4x_i^2+10x_i+4} > 0.46(n-1)$$ which disproves your conjecture.
For example, for $x_1=x_2=x_3=0.1$ and $x_4=1000$ we have $$\sum_{i=1}^4\sqrt\frac{x_i+1}{4x_i^2+10x_i+4}\approx1.417>\frac{4}{3}$$
The upper bound for $\sqrt\frac{x_i+1}{4x_i^2+10x_i+4}$ seems to approach 0.5, so we can probably say that for every $\epsilon>0$ there exists $(x_1,x_2,...,x_n)$ such that: $$\sum_{i=1}^n\sqrt\frac{x_i+1}{4x_i^2+10x_i+4} > \sum_{i=1}^{n-1}\sqrt\frac{x_i+1}{4x_i^2+10x_i+4} > (0.5-\epsilon)(n-1)$$ but I'm not qualified enough to go there.