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..

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

2 Answers 2


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!

  • $\begingroup$ 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). $\endgroup$ Jun 14, 2020 at 21:35
  • $\begingroup$ @Vendetta In this case I made it by hand and I am ready to explain. Sometimes I use WA. $\endgroup$ Jun 15, 2020 at 4:43
  • $\begingroup$ @MichaelRozenberg Why do we need to shift to $a, b, c$ again to show $k=1$ is indeed maximal? $\endgroup$ Jun 15, 2020 at 8:02
  • $\begingroup$ @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. $\endgroup$ Jun 15, 2020 at 10:47
  • $\begingroup$ Oh, I see. Thanks $\endgroup$ 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.


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