If $a,b,c\gt 0$, prove that $\frac {a^2}{b^3 c}- \frac {a}{b^2} \ge \frac{c}{b} - \frac{c^2}{a}$

Question

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?

Answer 1

Multiply both sides by $a/c^2$, and write the result as

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

Now let $x={a\over bc}$ and note that

$$x^3-x^2\ge x-1\iff x^3-x^2-x+1\ge0\iff(x-1)^2(x+1)\ge0\iff x+1\ge0$$

Since $a$, $b$, and $c$ are all assume to be positive, $x\gt0$, which implies $x+1\ge0$, so the inequality is proved.

The equality clearly holds when $x=1$ (i.e., when $x-1=0$), which is to say when $a=bc$.

Answer 2

Hint : Consider the expression \begin{eqnarray*} (a-bc)^2(a+bc) \geq 0. \end{eqnarray*}