<s>[INST] 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)
 [/INST] 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!
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