Prove that $\sum_{cyc}\frac{a^2}{ca^2 + 2c^2} \ge 1$

Question

$a$, $b$ and $c$ are positives such that $ab + bc + ca = 3abc$. Prove that $$ \sum_{cyc}\frac{a^2}{ca^2 + 2c^2} \ge 1$$

Here's what I did. My stupidity has reached a spiritual level.

We have that $ab + bc + ca = 3abc \implies \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 3$.

$$\sum_{cyc}\dfrac{a^2}{ca^2 + 2c^2} = \sum_{cyc}\dfrac{1}{c}\left(1 - \dfrac{2c}{a^2 + 2c}\right) \ge \sum_{cyc}\dfrac{1}{c}\left(1 - \dfrac{2c}{2c + 2a - 1}\right)$$

$$ = \left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) - 2\sum_{cyc}\dfrac{1}{2c + 2a - 1} \ge 3 - \dfrac{2}{9}\sum_{cyc}\left(\dfrac{1}{c} + \dfrac{1}{a} + \dfrac{1}{c + a - 1}\right)$$

$$ = 3 - \dfrac{4}{9}\left(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}\right) - \dfrac{2}{9}\sum_{cyc}\dfrac{1}{c + a - 1} \ge 3 - \dfrac{4}{3} - \dfrac{1}{18}\sum_{cyc}\left(\dfrac{1}{c - \frac{1}{2}} + \dfrac{1}{a - \frac{1}{2}}\right)$$

$$ = \dfrac{5}{3} - \dfrac{2}{9}\left(\dfrac{1}{2a - 1} + \dfrac{1}{2b - 1} + \dfrac{1}{2c -1}\right)$$

I am done with my life.

Answer 1

Let $\frac{1}{a}=x$, $\frac{1}{b}=y$ and $\frac{1}{c}=z$.

Thus, $x+y+z=3$ and by C-S we obtain: $$\sum_{cyc}\frac{a^2}{ca^2+2c^2}=\sum_{cyc}\frac{z^2}{2x^2+z}=\sum_{cyc}\frac{z^4}{2x^2z^2+z^3}\geq\frac{(x^2+y^2+z^2)^2}{\sum\limits_{cyc}(2x^2y^2+x^3)}.$$ Id est, it's enough to prove that $$(x^2+y^2+z^2)^2\geq\sum\limits_{cyc}(2x^2y^2+x^3)$$ or $$x^4+y^4+z^4\geq x^3+y^3+z^3.$$ Can you end it now?