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

$$a$$, $$b$$ 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.

## marked as duplicate by Shailesh, uniquesolution, GNUSupporter 8964民主女神 地下教會, Song, YiFanMar 24 at 13:26

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?
• @user574848 No. It's just Chebyshov. Also, Muirhead helps and there are more ways to the proof.For example: $\sum\limits_{cyc}(x^4-x^3)=\sum\limits_{cyc}(x^4-x^3-x+1)=\sum\limits_{cyc}(x-1)^2(x^2+x+1)\geq0.$ – Michael Rozenberg Mar 24 at 7:39