2. Positive numbers $a, b, c$ are such that $a+b+c=3$. Find the minimum value of the expression

$$
A=\frac{a^{3}+b^{3}}{8 a b+9-c^{2}}+\frac{b^{3}+c^{3}}{8 b c+9-a^{2}}+\frac{c^{3}+a^{3}}{8 c a+9-b^{2}}
$$
Answer: $\frac{3}{8}$.

Solution. Note that

$$
4\left(a^{3}+b^{3}\right) \geqslant(a+b)^{3} \quad \text { and } \quad 9-c^{2}=(3-c)(3+c)=(a+b)(3+c) .
$$

Then, by the inequalities for means,

$$
\frac{4\left(a^{3}+b^{3}\right)}{8 a b+9-c^{2}} \geqslant \frac{(a+b)^{3}}{2(a+b)^{2}+9-c^{2}}=\frac{(a+b)^{2}}{2(a+b)+3+c}=\frac{(3-c)^{2}}{9-c}=-c-3+\frac{36}{9-c}
$$

Similarly, the other two terms in $4A$ are estimated. Adding these inequalities, we get

$$
A \geqslant \frac{1}{4}\left(-(a+b+c)-9+\frac{36}{9-a}+\frac{36}{9-b}+\frac{36}{9-c}\right) \geqslant-3+\frac{81}{27-(a+b+c)}=\frac{3}{8}
$$

Equality is achieved when $a=b=c=1$.