<s>[INST] 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..
 [/INST] 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!
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