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) $ Find maximum $k  \in \mathbb{R}^{+}$ such that $$ \frac{a^3}{(b-c)^2} + \frac{b^3}{(c-a)^2} + \frac{c^3}{(a-b)^2} \geq k (a+b+c) $$
for all $a, b, c$ that are distinct positive real numbers ( $a \neq b$,  $b \neq c$,  $a \neq c$)
Usually when I see this kind of cyclic, symmetrical inequality, the extreme values are taken at $a = b = c$， which is obviously not the case here. So I am not sure how to approach this one..

This is a partial answer when $a,b,c$ form a side of a triangle.
Letting $c\to 0^+$, the inequality is in the form of:
$$\dfrac{a^3}{b^2}+\dfrac{b^3}{a^2}-k(a+b)\geq o(c),$$
for $o(c)\geq 0.$
But the left hand side is:
$$(a+b)\left(\dfrac{(a-b)^2(a^2+ab+b^2)}{a^2b^2}\right)+(a+b)(1-k).$$
If $k>1,$ then the whole thing can be made negative by taking $a = n+\epsilon$ and $b = n:$
$$(2n+\epsilon)\left(\epsilon^2\cdot\dfrac{n^2+3n\epsilon+3\epsilon^2}{n^2(n+\epsilon)^2} + 1-k\right).$$
For $k=1,$ the inequality is equivalent to:
$$\sum\dfrac{a(a-b+c)(a+b-c)}{(b-c)^2}\geq 0.$$
This begs for the Ravi substitution: $a = y+z,...$ and so:
$$\sum\dfrac{yz(y+z)}{(y-z)^2}\geq 0\iff \sum yz(y+z)(x-y)^2(x-z)^2\geq 0.$$
But this one is a direct consequence of a generalized Schur or Vornicu-Schur inequality, which can be found in Theorem $4$ here.
For when $a,b,c$ do not form a triangle, exactly one of $x,y,z$ is negative and the Vornicu-Schur does not apply here, at least not directly. I have a feeling that this is already a solved problem in Vasc's famous inequality book if you happen to have access to it.
