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.