Solution. Suppose that $a+b+c=3$. The problem becomes
$$\sum_{c y c} \frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}} \leq 4$$

Notice that
$$\begin{aligned} & \frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}}-\frac{4}{3}=\frac{(a-1)\left(-4 a^{2}-15 a+27\right)}{4 a^{3}+(3-a)^{3}} \\ = & (a-1)\left(\frac{2}{3}+\frac{(a-1)\left(-2 a^{2}-12 a-9\right)}{4 a^{3}+(3-a)^{3}}\right) \leq \frac{2(a-1)}{3} . \end{aligned}$$

We conclude that
$$\sum_{c y c} \frac{(3+a)^{2}}{4 a^{3}+(3-a)^{3}} \leq \sum_{c y c}\left(\frac{4}{3}+\frac{2(a-1)}{3}\right)=4$$
$\nabla$
Example 8.3.6. Let $a, b, c, d$ be non-negative real numbers. Prove that
$$\frac{a}{b^{2}+c^{2}+d^{2}}+\frac{b}{c^{2}+d^{2}+a^{2}}+\frac{c}{d^{2}+a^{2}+b^{2}}+\frac{d}{a^{2}+b^{2}+c^{2}} \geq \frac{3 \sqrt{3}}{2} \cdot \frac{1}{\sqrt{a^{2}+b^{2}+c^{2}+d^{2}}}$$