Let be a,b,c,d non negative real numbers. Prove that :
$$\frac{a}{a^{2}+b^{2}+2}+\frac{b}{b^{2}+c^{2}+2}+\frac{c}{c^{2}+d^{2}+2}+\frac{d}{d^{2}+a^{2}+2}\le 1$$
I tried many attempts but still can't find the result. Attempt 1 : $$\sum_{cyc}\frac{a}{a^2+b^2+2}\leq\frac{1}{2}\sum_{cyc}\frac{a}{ab+1}\leq\frac{1}{4}\sum_{cyc}\sqrt{\frac{a}{b}}$$ but the final term is superior to 1
Attempt 2 : $$\sum_{cyc}\frac{a}{a^2+1+b^2+1}\leq\frac{1}{2}\sum_{cyc}\frac{a}{a+b}=2-\frac{1}{2}\sum_{cyc}\frac{b}{a+b}=2-\frac{1}{2}\sum_{cyc}\frac{1}{1+\frac{a}{b}}.$$ So I need to prove that: $$\sum_{cyc}\frac{1}{1+\frac{a}{b}}\geq2$$ so taking $x=\frac{a}{b}$$y=\frac{b}{c}$ $z=\frac{c}{d}$ $t=\frac{d}{a}$ with $xyzt=1$
The inequality I need to prove is $$\sum_{cyc}\frac{1}{1+x}\geq2$$ I tried using Jensen inequality to $f(x)=\frac{1}{1+x}$ but it doesn't work.
By AM-GM twice we obtain: $$\sum_{cyc}\frac{a}{a^2+b^2+2}=\sum_{cyc}\frac{a}{a^2+1+b^2+1}\leq\frac{1}{2}\sum_{cyc}\frac{a}{\sqrt{(a^2+1)(b^2+1)}}=$$ $$=\frac{1}{2}\sum_{cyc}\sqrt{\frac{a^2}{a^2+1}\cdot\frac{1}{b^2+1}}\leq\frac{1}{4}\sum_{cyc}\left(\frac{a^2}{a^2+1}+\frac{1}{b^2+1}\right)=1.$$