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.

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 $\ge 0$. 
