10. (20 points) Let $\lambda$ be a positive real number. For any pairwise distinct positive real numbers $a, b, c$, we have
$$
\frac{a^{3}}{(b-c)^{2}}+\frac{b^{3}}{(c-a)^{2}}+\frac{c^{3}}{(a-b)^{2}} \geqslant \lambda(a+b+c) \text {. }
$$

Find the maximum value of $\lambda$.
10. Let $a=\frac{1}{2}, b=\frac{1}{2}-\varepsilon, c=\varepsilon\left(0<b<c$. We can set
$$
a=c+x, b=c+y(x>y>0) \text {. }
$$

Then the left side of equation (1)
$$
\begin{array}{l}
=\frac{(c+x)^{3}}{y^{2}}+\frac{(c+y)^{3}}{x^{2}}+\frac{c^{3}}{(x-y)^{2}} \\
=\frac{x^{3}+3 x^{2} c+3 x c^{2}+c^{3}}{y^{2}}+ \\
\frac{y^{3}+3 y^{2} c+3 y c^{2}+c^{3}}{x^{2}}+\frac{c^{3}}{(x-y)^{2}} \\
> \frac{x^{3}+3 x^{2} c}{y^{2}}+\frac{y^{3}+3 y^{2} c}{x^{2}} \\
=\left(\frac{x^{3}}{y^{2}}+\frac{y^{3}}{x^{2}}\right)+3 c\left(\frac{x^{2}}{y^{2}}+\frac{y^{2}}{x^{2}}\right) \\
\geqslant \frac{(x+y)^{3}}{(x+y)^{2}}+3 \times 2 c \sqrt{\frac{x^{2}}{y^{2}} \cdot \frac{y^{2}}{x^{2}}} \\
= x+y+6 c \\
>x+y+3 c \\
= a+b+c .
\end{array}
$$

Thus, the conclusion holds when $\lambda=1$.
In summary, the maximum value of $\lambda$ is 1.