Prove ${\frac {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{ \frac {1+{c}^{3}}{1+c{a}^{2}}}\ge 3 $ 
Let $a,b,c \ge0$, prove the on equality: $${\frac
 {1+{a}^{3}}{1+a{b}^{2}}}+{\frac {1+{b}^{3}}{1+b{c}^{2}}}+{  \frac
 {1+{c}^{3}}{1+c{a}^{2}}}\ge 3  $$

I tried: $$LHS = \sum\frac 1{1+ab^2}+\sum \frac {a^4}{a+a^2b^2} \ge\frac 9{3+\sum ab^2} + \frac {(a^2+b^2+c^2)^2}{\sum a+ \sum (ab)^2}\ge...$$
but it seem like useless.
Look like better inequality is $ \left( 1+{a}^{3} \right)  \left( 1+{b}^{3} \right)  \left( 1+{c}^{3}
 \right) \ge \left( 1+a{b}^{2} \right)  \left( 1+b{c}^{2} \right) 
 \left( 1+c{a}^{2} \right)$ (but still can't prove it)
 A: Without loss of generality, let $a\geqslant b\geqslant c\geqslant0$. The target inequality is equivalent to: $$\left(\frac{1+{a}^{3}}{1+a{b}^{2}}-1\right)+\left(\frac{1+{b}^{3}}{1+b{c}^{2}}-1\right)+\left(\frac{1+{c}^{3}}{1+c{a}^{2}}-1\right)\geqslant0$$
That is 
\begin{align*}
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\frac{c^2-a^2}{1+ca^2}\geqslant0\Leftrightarrow\\
a\frac{a^2-b^2}{1+ab^2}+b\frac{b^2-c^2}{1+bc^2}+c\left(\frac{c^2-b^2}{1+ca^2}+\frac{b^2-a^2}{1+ca^2}\right)\geqslant0\Leftrightarrow\\
(a^2-b^2)\left(\frac{a}{1+ab^2}-\frac{c}{1+ca^2}\right)+(b^2-c^2)\left(\frac{b}{1+bc^2}-\frac{c}{1+ca^2}\right)\geqslant0
\end{align*}
Think of the numerators of $\frac{a}{1+ab^2}-\frac{c}{1+ca^2}$ and $\frac{b}{1+bc^2}-\frac{c}{1+ca^2}$ are $$(a-c)+ac(a^2-b^2)\geqslant0~\mbox{and}~(b-c)+bc(a^2-c^2)\geqslant0$$Hence we complete the proof.
A: By AM-GM and Holder we obtain:
$$\sum_{cyc}\frac{1+a^3}{1+ab^2}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(1+a^3)}{\prod\limits_{cyc}(1+ab^2)}}=3\sqrt[3]{\frac{\sqrt[3]{\prod\limits_{cyc}(1+a^3)^3}}{\prod\limits_{cyc}(1+ab^2)}}=$$
$$=3\sqrt[3]{\frac{\sqrt[3]{\prod\limits_{cyc}(1+a^3)(1+b^3)^2}}{\prod\limits_{cyc}(1+ab^2)}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(1+ab^2)}{\prod\limits_{cyc}(1+ab^2)}}=3.$$
Done!
