Prove the inequality: $\dfrac{a}{1+b^{2}c}+\dfrac{b}{1+c^{2}d}+\dfrac{c}{1+d^{2}a}+\dfrac{d}{1+a^{2}b}\geq 2$

Question

Given $a+b+c+d = 4$

Prove the inequality

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

My Attempt:

$\dfrac{a^{2}}{a+ab^{2}c}+\dfrac{b^{2}}{b+bc^{2}d}+\dfrac{c^{2}}{c+cd^{2}a}+\dfrac{d^{2}}{d+da^{2}b}\geq \dfrac{\left( a+b+c+d\right) ^{2}}{a+b+c+d+ab^{2}c+bc^{2}d+cd^{2}a+da^{2}b}$

If the Right Hand Side is bigger than 2 than it is proves the former

$\dfrac{16}{4+\sum _{cyc}ab^{2}c}\geq 2$

$4\geq \sum _{cyc}ab^{2}c$

This is where I am stuck.