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.