## Prove that $\sum_{cyc}\dfrac{a}{b^2} \ge 3 \cdot \sum_{cyc}\dfrac{1}{a^2}$.

$$a$$, $$b$$ and $$c$$ are three positives such that $$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$$. Prove that $$\large \dfrac{a}{b^2} + \dfrac{b}{c^2} + \dfrac{c}{a^2} \ge 3 \cdot \left(\dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{1}{c^2}\right)$$

Here’s what I did.

We have that $$\left(\dfrac{a}{b^2} + \dfrac{b}{c^2} + \dfrac{c}{a^2}\right)\left(\dfrac{1}{b} + \dfrac{1}{c} + \dfrac{1}{a}\right) \ge \left(\sqrt{\dfrac{a}{b^3}} + \sqrt{\dfrac{b}{c^3}} + \sqrt{\dfrac{c}{a^3}}\right)^2$$

But because of $$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 1$$.

$$\implies \dfrac{a}{b^2} + \dfrac{b}{c^2} + \dfrac{c}{a^2} \ge 3 \cdot \left(\dfrac{1}{b}\sqrt{\dfrac{a}{c^3}} + \dfrac{1}{c}\sqrt{\dfrac{b}{a^2}} + \dfrac{1}{a}\sqrt{\dfrac{c}{b^2}}\right)$$

And I am stuck, I can’t think anymore.