$\sum_{cyc}\frac{a^3}{1+9b^2ac}=\sum_{cyc}\frac{a^4}{a+9a^2b^2c}\geq\frac{(a^2+b^2+c^2)^2}{a+b+c+9abc}\geq\frac{a+b+c}{6}>\frac{a+b+c}{18}$.

Dear arqady solved the problem correctly, but after solving dear shohvanilu changed the problem. Now I'm solving the new problem. As proven: $ \sum_{cyc}\frac{a^{3}}{1+9b^{2}ac}\geq \frac{(a^{2}+b^{2}+c^{2})^{2}}{a+b+c+9abc}$ So we just need to prove that $\frac{(a^{2}+b^{2}+c^{2})^{2}}{a+b+c+9abc} \geq \frac{(a+b+c)^{3}}{18} $ I have two solutions for this part (I'm posting all of my solutions just because it was an NMO not just a question): First solution using AM-GM: By Cauchy-Schwartz we have: $3(a^2+b^2+c^2)\geq (a+b+c)^2$ So we just need to prove: $\frac{(a+b+c)^4}{a+b+c+9abc}\geq \frac{(a+b+c)^3}{2}\Leftrightarrow a+b+c\geq 9abc\Leftrightarrow (a+b+c)(ab+bc+ca)\geq 9abc\Leftrightarrow a^2 b+a^2 c+a b^2+a c^2+b^2 c+b c^2\geq 6 a b c$ Which is obviously true by AM-GM.$\blacksquare$ Second solution using UVW Method: We have $3v^2=1\implies v=\frac{1}{\sqrt{3}}$. And also $3u=a+b+c\:,\:w^3=abc$ and $u\geq v\geq w$. So we need to prove that: $\frac{(9u^2-2)^2}{3u+9w^3}\geq \frac{27u^3}{18}$ After expanding all and do a little calculations we just need to prove: $1377u^4-243u^3w^3-648u^2+72\geq 0\Leftrightarrow 153u^4-27u^3w^3-72u^2+8\geq 0$ We know that $w\leq \frac{1}{\sqrt{3}}\implies -w^3\geq -\frac{1}{3\sqrt{3}}$ so by using this inequality we just need to prove: $153u^4-3\sqrt{3}u^3-72u^2+8\geq 0$ Define $f(u)=153u^4-3\sqrt{3}u^3-72u^2+8$ and we want to prove that $\forall u\geq \frac{1}{\sqrt{3}}$ we have $f(u)\geq 0$. $f'(u)=612 u^3-9 \sqrt{3} u^2-144 u=9 u (68 u^2-\sqrt{3} u-16)\implies \text{Extreme Poins of }f \text{ are:}\left\{\begin{matrix} u=0\\ u\approx -0.472503\\ u\approx 0.497974 \end{matrix}\right.$ We know that $0.497974< \frac{1}{2}\:\text{ and }f(\frac{1}{2})\approx -1.08702< 0$ and also $f(\frac{1}{\sqrt{3}})=0$ implies that $f$ is strictly increasing $\forall u\geq 0.497974$, So $\forall u\geq \frac{1}{\sqrt{3}}$ we have $f(u)\geq f(\frac{1}{\sqrt{3}}) \geq 0$ and we have equality case when $u=\frac{1}{\sqrt{3}}$ means $a=b=c=\frac{1}{\sqrt{3}}$.$\blacksquare$ Third solution making homogenizing: We can use $ab+bc+ca=1$ to make the last inequality homogeneous: $\frac{(a^{2}+b^{2}+c^{2})^{2}}{a+b+c+9abc} \geq \frac{(a+b+c)^{3}}{18} \Leftrightarrow \frac{(a^{2}+b^{2}+c^{2})^{2}(ab+bc+ca)}{(a+b+c)(ab+bc+ca)+9abc}\geq \frac{(a+b+c)^{3}}{18}\Leftrightarrow 17\sum_{sym}a^5b-4\sum_{sym}a^4b^2+30\sum_{cyc}a^3b^3-13abc\sum_{sym}a^2b-90a^2b^2c^2\geq 0$ $\text{Muirhead}([5,1,0]\geq [4,2,0])\implies 4\sum_{sym}a^5b\geq 4\sum_{sym}a^4b^2$ $\text{Muirhead}([5,1,0]\geq [3,2,1])\implies 13\sum_{sym}a^5b\geq 13\sum_{sym}a^3b^2c=13abc\sum_{sym}a^2b$ $\text{AM-GM}\implies 30\sum_{sym}a^3b^3\geq 90a^2b^2c^2$ By summing up these three inequalities the main inequality would be desired.$\blacksquare$

Use Holder $(\sum_{cyc} 1) (\sum_{cyc} (1+9b^2 ac) (\sum_{cyc} \frac{a^3} {1+9b^2 ac}) \geq (\sum_{cyc} a)^3$ and this basic inequality, it's done. $1=(ab+bc+ca)^2 \geq 3abc(a+b+c)$

Given $a,b$ and $c$ positive real numbers with $ab+bc+ca=1$. Then prove that $$\frac{a^3}{1+9b^2ac}+\frac{b^3}{1+9c^2ab}+\frac{c^3}{1+9a^2bc} \geq \frac{a^3+b^3+c^3}{2}$$.

Notice that $\sum_{cyc}\frac{a^3}{1+9b^2ac}\overset{\text{Generalized T2’S}}{\ge}\frac{1}{3}\cdot\frac{\left(\sum_{cyc}a\right)^3}{3+9abc\sum_{cyc}a}$ Thus the inequality transforms into $\frac{1}{3}\cdot\frac{\left(\sum_{cyc}a\right)^3}{3+9abc\sum_{cyc}a}\ge\frac{\left(\sum_{cyc}a\right)^3}{18}\Longleftrightarrow\frac{1}{3+9abc\sum_{cyc}a}\ge\frac{1}{9}$ Furthermore $9\ge3+9abc\sum_{cyc}a\Longleftrightarrow1\ge abc\sum_{cyc}a$ Which is clearly true as $1=\left(\sum_{cyc}ab\right)^2\ge abc\sum_{cyc}a$ $\blacksquare$.