If $a,b,c\gt 0$,

$1º$ Prove that:

$$\frac {a^2}{b^3 c}- \frac {a}{b^2} \ge \frac{c}{b} - \frac{c^2}{a}$$

$2º$ For which values of $a,b,c$ the equality holds?

My try:

I tried a lot of things, but none of them worked (example):

$$\frac {a^2}{b^3 c}- \frac {a}{b^2} \ge \frac{c}{b} - \frac{c^2}{a}$$
$$\frac {a}{b^2} \left(\frac {a}{bc}-1\right) \ge c \left( \frac{1}{b} - \frac{c}{a} \right)$$
$$\frac {a^2b}{b^2} \left(\frac {a-bc}{bc}\right) \ge c \left( {a-bc} \right)$$
$$ a^2 \left(a-bc \right) \ge b^2c^2 \left( a-bc \right)$$
$$ a^2\ge b^2c^2$$

After this, i don't know how to proceed, and maybe it's the wrong way.

Any hints?