$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq 2$

Question

I am stuck in the following olympiad problem:

Suppose that $a,b,c,d\geq 0$ and $a+b+c+d=4$. Prove that $$ \frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq 2. $$

Attempt. I tried to use reversed AM-GM technique. \begin{alignat}{2} \frac{a}{1+b^2c}-a &= \frac{-ab^2c}{1+b^2c} &\qquad \frac{b}{1+c^2d}-b &= \frac{-bc^2d}{1+c^2d} \\ \frac{c}{1+d^2a}-c &= \frac{-cd^2a}{1+d^2a} &\qquad \frac{d}{1+a^2b}-d &= \frac{-da^2b}{1+a^2b} \end{alignat}

Answer 1

By C-S and AM-GM we obtain: $$\sum_{cyc}\frac{d}{1+a^2b}=\sum_{cyc}\frac{d^2}{d+a^2bd}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(d+a^2bd)}=$$ $$=\frac{16}{4+(ab+cd)(ad+bc)}\geq\frac{16}{4+\left(\frac{ab+bc+cd+da}{2}\right)^2}=$$ $$=\frac{16}{4+\left(\frac{(a+c)(b+d)}{2}\right)^2}\geq\frac{16}{4+\left(\frac{\left(\frac{a+c+b+d}{2}\right)^2}{2}\right)^2}=2.$$

Answer 2

$$\sum_{cyc}\frac {d}{1+a^2b}=\sum_{cyc}\frac {d^2}{d+a^2bd}$$

Hence by Titu's lemma $$\sum_{cyc}\frac {d^2}{d+a^2bd}\ge \frac {(a+b+c+d)^2}{\sum_{cyc} (d+a^2bd)}=$$ $$=\frac{16}{4+(ab+cd)(ad+bc)}$$

Now by AM GM inequality we have $$\frac {ab+bc+cd+ad}{2}\ge \sqrt{(ab+cd)(ad+bc)}$$ Hence we get $$\left(\frac {ab+bc+cd+ad}{2}\right) ^2\ge (ab+cd)(ad+bc)$$

$$\Rightarrow \frac{16}{4+(ab+cd)(ad+bc)} \ge \frac {16}{4+\left(\frac {ab+bc+cd+ad}{2}\right)^2}$$

Now again by AM GM inequality $$\frac {a+b+c+d}{2}\ge \sqrt{(a+c)(d+b)}$$

$$\Rightarrow \frac {16}{4+\left(\frac {(a+c)(b+d)}{2}\right)^2}\ge \frac {16}{4+\left(\frac {\left(\frac {a+b+c+d}{2}\right)^2}{2}\right)^2}$$ $$=2.$$