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.
 A: Write 
$$a = \frac{p^3}{p q r},\, b= \frac{q^3}{p q r}, \,
c= \frac{r^3}{p q r}
$$
with $0 < p\le q \le r$ and substitute to get the equivalent inequality:
$$
p q^8 r^4 + p^5 q^3 r^5 + q r^{12}-p^6 q^7   - p q^5 r^7 - p^5 r^8 \ge 0$$
Now write $p=u,\, q=u+v, \,r=u+v+w$, with $u>0$, $\ v,w\ge 0$ and substitute to get an expression in $u$,$v$,$w$ with all coefficients positive, hence $\ge 0$. 
A: 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$=a-\dfrac{1}{a}+b^2-\dfrac{1}{b^2}+c^3-\dfrac{1}{c^3} \ge a-\dfrac{1}{a}+c^3-\dfrac{1}{c^3} \ge c^3-c^2+\dfrac{1}{c^2}-\dfrac{1}{c^3}=\dfrac{c^2(c-1)(c^5+1)}{c^5} \ge 0$
A: we can write the inequality as
$$\left(a-\dfrac{1}{c^3}\right)+\left(c^3-\dfrac{1}{a}\right)\ge\dfrac{1}{b^2}-b^2$$
or
$$(c^3a-1)\left(\dfrac{1}{a}+\dfrac{1}{c^3}\right)\ge\dfrac{1-b^4}{b^2}\tag{1}$$
since $c\ge 1,bc\ge 1$,then
$$c^3a\ge c\cdot abc=c\ge 1$$
so it only prove $b\le 1$case.
$1$.case if $b^2\ge a$，then we  have
$$c^3a+b^4-2\ge a(c^3+b^3)-2\ge abc(b+c)-2=b+c-2\ge 0$$
then we have
$$c^3a-1\ge 1-b^4$$
and
$$\dfrac{1}{a}\ge\dfrac{1}{b^2}$$
so $(1)$ prove done.
2. if $a\ge b^2$,we have
$$c^3a-1-c(1-b^4)\ge c^3b^2+cb^4-c-1\ge c^3b^2+cb^4-c(b^2c)^{4/3}-(b^2c)^{5/3}\ge 0$$
so we have
$$c^3a-1\ge c(1-b^4)$$
and
$$\dfrac{c}{a}-\dfrac{1}{b^2}\ge\dfrac{b-a}{ab^2}\ge 0$$
so
$$(c^3a-1)\left(\dfrac{c}{a}+\dfrac{1}{c^2}\right)\ge\dfrac{c(1-b^4)}{b^2}$$
so $(1)$ prove by done
