# 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..

• @jeff123 that's why I said I dont know how to approach this one.. Jun 14, 2020 at 14:20
• "Where $a,b,c$ are distinct positive real numbers" is ambiguous. Better is "whenever $a,b,c$ are distinct positive real numbers." Jun 14, 2020 at 14:26

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!

• The last step doesn't seem so trivial though. Out of curiosity are you always able to manually refactor the polynomials or you use computers (nothing against that, just curious). Jun 14, 2020 at 21:35
• @Vendetta In this case I made it by hand and I am ready to explain. Sometimes I use WA. Jun 15, 2020 at 4:43
• @MichaelRozenberg Why do we need to shift to $a, b, c$ again to show $k=1$ is indeed maximal? Jun 15, 2020 at 8:02
• @Shiv Tavker We need to show that for $k=1$ our inequality is true for any real and different variables. Since we saw that $k\leq1$, it says that $1$ is a maximal value of $k$, for which it happens. Jun 15, 2020 at 10:47
• Oh, I see. Thanks Jun 15, 2020 at 11:31

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.