3.5.10 ** Find the largest real number $k$ such that the inequality
$$
(a+b+c)^{3} \geqslant \frac{5}{2}\left(a^{3}+b^{3}+c^{3}\right)+k a b c
$$

holds for the side lengths $a, b, c$ of any triangle.
When $a=b=c$, we have $k \leqslant \frac{39}{2}$.
Below, we prove the inequality $(a+b+c)^{3} \geqslant \frac{5}{2}\left(a^{3}+b^{3}+c^{3}\right)+\frac{39}{2} a b c$
holds for any side lengths $a, b, c$ of a triangle.
(1) $\Leftrightarrow(3 c-a-b)(a-b)^{2}+(3 a-b-c)(b-c)^{2}+(3 b-c-a)(c-a)^{2} \geqslant 0$.

Since the sum of any two of $3 c-a-b, 3 a-b-c, 3 b-c-a$ is positive, at most one of these numbers can be negative.
(1) If none of $3 c-a-b, 3 a-b-c, 3 b-c-a$ are negative, then (2) holds.
(2) If exactly one of $3 c-a-b, 3 a-b-c, 3 b-c-a$ is negative, assume without loss of generality that $3 c-a-b$ is negative, i.e., $3 c-a-b < 0$, while $3 a-b-c \geqslant 0$ and $3 b-c-a \geqslant 0$. Since $3 b-c-a > a+b-3 c > 0$, we have $(3 b-c-a)(a-c)^{2} > (a+b-3 c)(a-b)^{2}$, thus (2) holds.
In conclusion, the maximum value of $k$ is $\frac{39}{2}$.