<s>[INST] This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$ 
Let $0\le a\le b\le c,abc=1$, then show that
  $$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
  Things I have tried so far:
  $$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$

Since $abc=1$, it suffices to prove that
$$c+b^3c^2+bc^5\ge b^2c^3+ac^3+ab^2$$
then the problem is solved. I stuck in here.
 [/INST] hint: $c\ge 1, a \le 1$
when $b\le 1 $. replace $a$,get $f(c)=$LHS-RHS,prove $f'(c)\ge0$ with 
$ bc=\dfrac{1}{a} \ge 1,$ then use $ c\ge \dfrac{1}{b^2} $ to prove $f(c=\dfrac{1}{b^2} )\ge 0$
Edit: here is full solution:
case 1: $b \le 1, c \ge \dfrac{1}{b^2},bc=\dfrac{1}{a} \ge 1$
LHS-RHS$=b^2c^6-b^3c^4+b^4c^3-c^3+bc^2-b^2=f(c),\\f'(c)=6b^2c^5-4b^3c^3+3b^4c^2-3c^2+2bc=4b^2c^3(c^2-b)+2b^2c^5-3c^2+2bc+ 3b^4c^2 >0 $
becasue $c^2 \ge1 \ge b, 2b^2c^5+2bc-3c^2 > 2c^3-3c^2+1=(c-1)(2c^2-c-1)\ge 0 $ 
$f(c)_{min}=f(\dfrac{1}{b^2})=\dfrac{(1-b^3)(b^9+b^6-b^5-b^4+b^3+1 )}{b^{10}} \ge 0$
when $a=b=1$ get min $0$
case 2: $b \ge 1, \implies b^2 -\dfrac{1}{b^2} \ge 0,a=\dfrac{1}{bc} \ge \dfrac{1}{c^2}$
it is trivial $g(a)=a-\dfrac{1}{a}$ is mono increasing function,$\implies g(a) \ge \dfrac{1}{c^2} -c^2$
LHS-RHS$=a-\dfrac{1}{a}+b^2-\dfrac{1}{b^2}+c^3-\dfrac{1}{c^3} \ge a-\dfrac{1}{a}+c^3-\dfrac{1}{c^3} \ge c^3-c^2+\dfrac{1}{c^2}-\dfrac{1}{c^3}=\dfrac{c^2(c-1)(c^5+1)}{c^5} \ge 0$
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