$a, b, c$ are reals such that $a^2 + b^2 + c^2 = 1$. Calculate the maximum value of $$\large |ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)|$$

This problem appeared in a 9th-grade book about elementary ways to prove inequalities. And I struggled hard with this problem in particular. (I'm a student in 10th grade now.)

I'm still questioning if they should delete this example (not an actual exercise), (and they needn't, but the writers of the book should suggest this problem to the authors of a different book for further education in Mathematics).

Furthermore, there are problems which are marked difficult or challenging. This one doesn't (What are the attributes of a hard problem that this one doesn't have?).

It is unusual. In retrospect, I can't solve it.

We have that $$|ab(a^2 - b^2) + bc(b^2 - c^2) + ca(c^2 - a^2)| = |(a + b + c)(a - b)(b - c)(c - a)|$$

And I'm done, factorizing the expression is all I got.

The maximum value is $\dfrac{9\sqrt 2}{32}$, occurs when $(a, b, c) = \left(\dfrac{2 - 3\sqrt 2}{4\sqrt 3}, \dfrac{1}{2\sqrt 3}, \dfrac{2 + 3\sqrt 2}{4\sqrt 3}\right)$ and all of the permutations of the aforementioned.

Please help me.

I have tried using the AM-GM inequality, the Cauchy inequality, the rearrangement inequality, unidentified variables but I have found no success.

  • 1
    $\begingroup$ It's fourth degree. Use uvw $\endgroup$ – Michael Rozenberg Jun 29 at 16:47
  • $\begingroup$ It's s quadratic inequality of w^3 $\endgroup$ – Michael Rozenberg Jun 29 at 16:50
  • $\begingroup$ Actually this problem is a problem 3 of IMO 2006. See here artofproblemsolving.com/community/c6h101294p571945 $\endgroup$ – richrow Jun 29 at 16:58
  • $\begingroup$ Assume positive value for the absolute term and try partial derivatives. $\endgroup$ – Moti Jun 29 at 19:24

Let $k$ be a maximal value.

Thus, $k$ it's a minimal number, for which the inequality $$\left|\sum_{cyc}(a^3b-a^3c)\right|\leq k(a^2+b^2+c^2)^2$$ is true for all reals $a$, $b$ and $c$. We see that $k>0$.

Now, let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative, $abc=w^3$ and $u^2=tv^2$. Thus, $$|(a+b+c)(a-b)(a-c)(b-c)|\leq k(a^2+b^2+c^2)^2$$ or $$(a+b+c)^2\prod_{cyc}(a-b)^2\leq k^2(a^2+b^2+c^2)^4$$ or $$9u^2\cdot27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\leq81k^2(3u^2-2v^2)^4$$ or $$u^2w^6+2u^3(2u^2-3v^2)w^3-3u^4v^4+4u^2v^6+\frac{k^2}{3}(3u^2-2v^2)^4\geq0,$$ for which we need $$u^6(2u^2-3v^2)^2-u^2\left(-3u^4v^4+4u^2v^4+\frac{k^2}{3}(3u^2-2v^2)^4\right)\leq0$$ or $$t^2(2t-3)^2+3t^2-4t\leq\frac{k^2}{3}(3t-2)^4$$ or $$\frac{12t(t-1)^3}{(3t-2)^4}\leq k^2.$$ Consider two cases.

  1. $t>0.$

Thus, $t\geq1$ and in this case $$\frac{12t(t-1)^3}{(3t-2)^4}\leq\lim_{t\rightarrow+\infty}\frac{12t(t-1)^3}{(3t-2)^4}=\frac{4}{27}.$$ Indeed, $$\frac{4}{27}-\frac{12t(t-1)^3}{(3t-2)^4}=\frac{4(27t^3-27t^2-15t+16)}{27(3t-2)^4}=\frac{4((t-1)(27t^2-15)+1)}{27(3t-2)^4}>0.$$

  1. $t<0$.

Thus, by AM-GM $$\frac{12t(t-1)^3}{(3t-2)^4}=\frac{-3t(2-2t)^3}{2(3t-2)^4}\leq\frac{\left(\frac{-3t+3(2-2t)}{4}\right)^4}{2(3t-2)^4}=\frac{81}{512}$$ and since $$\frac{81}{512}>\frac{4}{27},$$ we obtain:$$k^2=\frac{81}{512}$$ and $$k=\frac{9}{16\sqrt2}.$$ The equality occurs for $$-3t=2-2t$$ or $$t=-2,$$ which is possible, which says that $\frac{9}{16\sqrt2}$ is the answer.


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