I am trying to us Cauchy-Schwarz inequality to find the maximum of: $$|(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)|$$ Where $a$, $b$, and $c$ are real numbers, and $a+b+c=0$ and $a^2+b^2+c^2=2$.
What I've tried:
$$a_1=a,\ a_2=b,\ a_3=c$$ $$b_1=b\sqrt{a-b}, \ b_2=c\sqrt{b-c}, \ b_3=a\sqrt{c-a}$$ Which gives: $$(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)\leq(a^2+b^2+c^2)(a-b^3+b-c^3+c-a^3)$$ And I can rearrange and apply the constraints to get $$(a^2)(b^2)(a-b)+(b^2)(c^2)(b-c)+(c^2)(a^2)(c-a)\leq2(-b^3-c^3-a^3)$$ But I'm not sure where to go from here. How can I use CS to get to an answer here?
Let $a=1$, $b=-1$ and $c=0$. Hence, we get a value $2$. We'll prove that it's a maximal value.
Indeed, since the condition gives $ab+ac+bc=-1$, we obtain:
$|\sum\limits_{cyc}(a^3b^2-a^3c^2)|=|(ab+ac+bc)(a-b)(a-c)(b-c)|=|(a-b)(2a+b)(a+2b)|$.
Since the condition gives $a^2+ab+b^2=1$, it remains to prove that $$4(a^2+ab+b^2)^3\geq(a-b)^2(2a^2+5ab+2b^2)^2$$
which is $(a+b)^2a^2b^2\geq0$. Done!