If $a,b,c\in R: a+b+c=3,$ find the maximum $$P=\min\{a,b,c\}+ \sqrt[3]{abc}-\max\{a,b,c\}. $$
Some thoughts.
I think we just need to prove $\sqrt[3]{abc}\le 1,$ which is true by AM-GM for $a,b,c\ge 0.$
I don't know how to prove in the remain case of $a,b,c.$
May be there is some good ideas. Hope to see some helps, thank you.
For $a=b=c=1$ we obtain a value $1$.
We'll prove that it's a maximal value.
Indeed let $a\geq b\geq c$.
Thus, $a\geq0$ and we need to prove that: $$c+\sqrt[3]{abc}-a\leq\frac{a+b+c}{3}$$ or $$4a+b-2c\geq3\sqrt[3]{abc},$$ which is obvious for $abc<0$ and $abc\geq0,$ where $a$, $b$ and $c$ are non-negatives.
Let $b\leq0$ and $c\leq0$.
Thus, by AM-GM $$4a+b-2c=3a+a-b+2(b-c)\geq 2a=$$ $$= a+(-b)+(-c)+a+b+c\geq $$ $$\geq a+(-b)+(-c)\geq3\sqrt[3]{a(-b)(-c)}=3\sqrt[3]{abc}.$$