Find maximum $k \in \mathbb{R}^{+}$ such that $ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $

Let $c\rightarrow0^+$.

Thus, $$\frac{a^3}{b^2}+\frac{b^3}{a^2}\geq k(a+b),$$ which gives that $k\leq1.$

We'll prove that $1$ is a maximal value.

Indeed, we need to prove that: $$\sum_{cyc}\frac{a^3}{(b-c)^2}\geq a+b+c.$$ Now, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.

Thus, we need to prove that: $$(u^2-uv+v^2)^2a^3+3(u^3+v^3)(u-v)^2a^2+3(u^4-u^2v^2+v^4)(u-v)^2a+$$ $$+(u+v)(u^2+uv+v^2)(u-v)^4\geq0$$ and we are done!