For $a,b,c$ are real number satisfied: $a^3+b^3+c^3=3abc+32$. Find minimize value of $P=(a^2+b^2+c^2)(|a-b|+|b-c|+|c-a|)$
This is my try: WLOG, I suppose that: $a\ge b\ge c$.
Then we have $P=(a^2+b^2+c^2)(a-b+b-c+a-c)=(a^2+b^2+c^2)(2a-2c)$
In the other hand, we also have $(b-a)(b-c)\le 0\iff b^2+ac\le ab+bc\iff b^3+abc\le ab^2+b^2c$.
In addition, we have $(a-b)(b-c)(c-a)\le 0\iff (ab^2+bc^2+ca^2)-(a^2b+b^2c+c^2a)\le 0$
I have tried exploiting the assumptions that the problem gave, but I can't link these to come final solution !
Let $a\geq b\geq c$.
Thus, since $(a-c)^2\geq\sum\limits_{cyc}(a^2-ab)$ it's $(a-b)(b-c)\geq0,$ by AM-GM we obtain: $$P=\sqrt{(a^2+b^2+c^2)^2(a-b+b-c+a-c)^2}=$$ $$=\frac{1}{3}\sqrt{\left((a+b+c)^2+\sum_{cyc}(a-b)^2\right)^2\cdot4(a-c)^2}\geq$$ $$\geq\frac{1}{3}\sqrt{4(a+b+c)^2\sum_{cyc}(a-b)^2\cdot4\sum_{cyc}(a^2-ab)}=$$ $$=\frac{1}{3}\sqrt{32\left((a+b+c)\sum_{cyc}(a^2-ab)\right)^2}=$$ $$=\frac{1}{3}\sqrt{32\left(a^3+b^3+c^3-3abc\right)^2}=\frac{1}{3}\sqrt{32^3}=\frac{128\sqrt2}{3}.$$ The equality occurs for $(a-b)(b-c)=0$, $(a+b+c)^2=\sum\limits_{cyc}(a-b)^2$ and $a^3+b^3+c^3-3abc=32,$ which says that we got a minimal value.