Let $a$, $b$ and $c$ be positive real numbers such that $abc \geq 10+6\sqrt{3}$. Prove that: $$ \displaystyle\sum_{cyc} \frac{a}{c+a^3+b^2} \leq \frac{1}{2}.$$
I've tried to bound the LHS of Michael Rozenberg's last inequality,
$\frac{3\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5(10+6\sqrt3)^{\frac{6}{16.5}}} = \frac{3}{16.5}\cdot\left(\frac{7.5^{7.5}\cdot27\cdot6^6}{(10+6\sqrt3)^6}\right)^{\frac{1}{16.5}} < \frac{3}{16.5}\cdot\left(\frac{7.5^{7.5}\cdot27\cdot6^6}{20^6}\right)^{\frac{1}{16.5}}$
$7.5^{7.5}=7.5^6\cdot7.5^{1.5}<7.5^6\cdot7.5^2 <7.5^6\cdot2^6$
$<\frac{3}{16.5}\cdot\left(\frac{15^6\cdot(\sqrt{3})^6\cdot6^6}{20^6}\right)^{\frac{1}{16.5}}=\frac{3}{16.5}\cdot\left(\frac{90\sqrt{3})^6}{20^6}\right)^{\frac{1}{16.5}}$
$=\frac{3}{16.5}\cdot\left(\frac{90\sqrt{3}}{20}\right)^{\frac{6}{16.5}}=\frac{3}{16.5}\cdot(4.5\sqrt{3})^{\frac{6}{16.5}}$
$<\frac{3}{16.5}\cdot(4.5\sqrt{3})^{\frac{1}{2}}<\frac{3}{16.5}\cdot(8)^{\frac{1}{2}}<\frac{3}{16.5}\cdot3=0.54$
Let $a=(1+\sqrt3)kx^2$, $b=(1+\sqrt3)ky^2$ and $c=(1+\sqrt3)kz^2$, where $k>0$ and $xyz=1$.
Hence, $10+6\sqrt3\leq abc=(10+6\sqrt3)k^3xyz$, which gives $k\geq1$ and $$\sum_{cyc}\frac{a}{a^3+b^2+c}=\sum_{cyc}\frac{x}{k^2(1+\sqrt3)^2x^3+(1+\sqrt3)ky^2+z}\leq$$ $$\leq\sum_{cyc}\frac{x}{(1+\sqrt3)k(2\sqrt{x^3z}+y^2)}\leq\sum_{cyc}\frac{x}{3x^3+2y^2+z}.$$ Thus, it remains to prove that $$\sum_{cyc}\frac{x}{3x^3+2y^2+z}\leq\frac{1}{2},$$ which is true, but it's very ugly.
In another hand, by AM-GM we obtain: $$\sum_{cyc}\frac{a}{a^3+b^2+c}=\sum_{cyc}\frac{a}{7.5\cdot\frac{a^3}{7.5}+3\cdot\frac{b^2}{3}+6\cdot\frac{c}{6}}\leq$$ $$\leq\sum_{cyc}\frac{a}{16.5\left(\frac{a^3}{7.5}\right)^{\frac{7.5}{16.5}}\cdot\left(\frac{b^6}{27}\right)^{\frac{1}{16.5}}\cdot\left(\frac{c^6}{6^6}\right)^{\frac{1}{16.5}}}=$$ $$=\frac{\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5}\sum_{cyc}\frac{a}{a^{\frac{22.5}{16.5}}b^{\frac{6}{16.5}}c^{\frac{6}{16.5}}}=\frac{\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5}\sum_{cyc}\frac{1}{a^{\frac{6}{16.5}}b^{\frac{6}{16.5}}c^{\frac{6}{16.5}}}=$$ $$=\frac{3\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5(abc)^{\frac{6}{16.5}}}\leq\frac{3\left(7.5^{7.5}\cdot27\cdot6^6\right)^{\frac{1}{16.5}}}{16.5(10+6\sqrt3)^{\frac{6}{16.5}}}=0.3555...\leq\frac{1}{2}.$$ Done!