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.

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
