Nice! By C-S we have that $LHS=\sum_{cyc} \frac{a^4}{a^3+3ab^2+3a^2b+2abc} \geqslant \frac{(a^2+b^2+c^2)^2}{\sum (a^3+3ab^2+3a^2b+2abc)}=\frac{(a^2+b^2+c^2)^2}{(a+b+c)^3} >^{?} \frac{1}{6(a^2+b^2+c^2)^2}$ But by AM-GM we have $\sqrt[3]{6} (a^2+b^2+c^2)>\sqrt{3}(a^2+b^2+c^2) \geqslant (a+b+c)\frac{a+b+c}{\sqrt{3}} \geqslant (a+b+c)\sqrt{ab+bc+ca}=a+b+c$ and $a^2+b^2+c^2 \geqslant ab+bc+ca=1$ So, $6(a^2+b^2+c^2)^4 \geqslant 6(a^2+b^2+c^2)^3>(a+b+c)^3$, qed.

By Cauchy-Schwarz $$\left( \sum_{cyc} \frac{a^3}{a^2+3b^2+3ab+2bc}\right)\left(\sum_{cyc} a(a^2+3b^2+3ab+2bc)\right)\geq \left(a^2+b^2+c^2\right)$$Therefore, it´s sufficient to prove that \begin{align} 6(a^2+b^2+c^2)^4 &> \sum_{cyc} a(a^2+3b^2+3ab+2bc)\nonumber \\ &= a^3+b^3+c^3+3\Bigl(ab^2+a^2b+bc^2+b^2c+ca^2+c^2a\Bigr)+6abc\nonumber \\ &= a^3+b^3+c^3+3\Bigl((a+b+c)(ab+bc+ca)-3abc\Bigr)+6abc\nonumber \\ &=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3(a+b+c)=(a+b+c)(a^2+b^2+c^2+2)\nonumber \end{align}As $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2+2$, we need to prove $$6\Bigl((a+b+c)^2-2\Bigr)^4>(a+b+c)^3$$But this is clear since $(a+b+c)^2\geq 3(ab+bc+ca)=3$ and $6=6\left(\sqrt{3}^2-2\right)^4>(\sqrt{3})^3$

Posting for storage. First of all, notice that $\sum_{cyc}\frac{a^3}{a^2+3b^2+3ab+2bc}=\sum_{cyc}\frac{a^4}{a^3+3ab^2+3a^2b+2abc}\overset{\text{Engel C-S}}{\ge}\frac{\left(\sum_{cyc}a^2\right)^2}{\sum_{cyc}a^3+3\sum_{sym}a^2b+6abc}=\frac{\left(\sum_{cyc}a^2\right)^2}{\left(\sum_{cyc}a\right)^3}$ Thus the inequality is equivalent to $\frac{\left(\sum_{cyc}a^2\right)^2}{\left(\sum_{cyc}a\right)^3}>\frac{1}{6(a^2+b^2+c^2)^2}\Longleftrightarrow 6\left(\sum_{cyc}a^2\right)^4>\left(\sum_{cyc}a\right)^3$ Furthermore notice that $6\left(\sum_{cyc}a^2\right)^4=6\left(\sum_{cyc}a^2\right)^3\cdot\sum_{cyc}a^2\ge6\left(\sum_{cyc}a^2\right)^3\sum_{cyc}ab=6\left(\sum_{cyc}a^2\right)^3$ Therefore the inequality boils down to $6\left(\sum_{cyc}a^2\right)^3>\left(\sum_{cyc}a\right)^3\Longleftrightarrow \sqrt[3]{6}\sum_{cyc}a^2>\sum_{cyc}a$ Furthermore, notice that by $\text{AM-GM or C-S}$ we obtain $\sqrt[3]{6}\sum_{cyc}a^2>\sqrt{3}\sum_{cyc}a^2\ge\frac{\left(\sum_{cyc}a\right)^2}{\sqrt{3}}\ge\sum_{cyc}a\sqrt{\sum_{cyc}ab}=\sum_{cyc}a$ Therefore $\sqrt[3]{6}\sum_{cyc}a^2>\sqrt{3}\sum_{cyc}a^2\ge\sum_{cyc}a$ $\blacksquare$.

jbaca wrote: Let $a,\ b$ and $c$ be positive real numbers such that $a b+b c+c a=1$. Show that $$ \frac{a^3}{a^2+3 b^2+3 a b+2 b c}+\frac{b^3}{b^2+3 c^2+3 b c+2 c a}+\frac{c^3}{c^2+3 a^2+3 c a+2 a b}>\frac{1}{6\left(a^2+b^2+c^2\right)^2} . $$

I´m the author of this problem, Miguel Ángel Hernández from the Dominican Republic. I hope you found my problem interesting. Here is my solution: