<s>[INST] Prove $\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2} \ge 2$ if $a+b+c+d=4$ 
if $a,b,c,d$ are positive real numbers that $a+b+c+d=4$,Prove:$$\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+d^2}+\frac{d}{1+a^2} \ge 2$$
  Additional info:I'm looking for solutions and hint that using Cauchy-Schwarz and AM-GM because I have background in them.

Things I have done: Using AM-GM $$LHS=\sum\limits_{cyc}\left(a-\frac{ab^2}{1+b^2}\right)\ge \sum\limits_{cyc}\left(a-\frac{ab^2}{2b}\right)=\sum\limits_{cyc}a-\frac{1}{2}(ab+bc+cd+da)$$
So it lefts to Prove that $$\sum\limits_{cyc}a-\frac{1}{2}(ab+bc+cd+da)\ge 2$$
If I show that $ab+bc+cd+da \le4$ then the upper inequality will be true.But I have problem in showing that.
 [/INST] You could continue using AM-GM
$$ab + bc + cd + da = (a+c)(b+d) \le \frac{(\overline{a+c}+\overline{b+d} )^2}4 = 4$$
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