Prove that $\sum_{cyc}\frac{1}{x^2 + 1} \ge \frac{2}{3}\bigl(\sum_{cyc}\frac{x}{\sqrt{x^2 + 1}}\bigr)^3$ where $x, y, z > 0$ and $xy + yz + zx = 1$.

Question

$x$, $y$ and $z$ are positives such that $xy + yz + zx = 1$. Prove that $$\large \frac{1}{x^2 + 1} + \frac{1}{y^2 + 1} + \frac{1}{z^2 + 1} \ge \frac{2}{3}\left(\frac{x}{\sqrt{x^2 + 1}} + \frac{y}{\sqrt{y^2 + 1}} + \frac{z}{\sqrt{z^2 + 1}}\right)^3$$

We have that $$\frac{x}{\sqrt{x^2 + 1}} = \frac{x}{\sqrt{x^2 + xy + yz + zx}} = \frac{x}{\sqrt{(x + y)(z + x)}} \le \frac{x}{x + \sqrt{yz}} = 1 - \frac{\sqrt{yz}}{x + \sqrt{yz}}$$

$$ \implies \sum_{cyc}\frac{x}{\sqrt{x^2 + 1}} \le 3 - \sum_{cyc}\frac{\sqrt{yz}}{x + \sqrt{yz}} = 3 - \sqrt{xyz}\sum_{cyc}\frac{1}{x\sqrt x + \sqrt{xyz}} \le 3\left(1 - \frac{3\sqrt{xyz}}{x\sqrt x + y\sqrt y + z\sqrt z + 3\sqrt{xyz}}\right) = \frac{3(x\sqrt x + y\sqrt y + z\sqrt z)}{x\sqrt x + y\sqrt y + z\sqrt z + 3\sqrt{xyz}}$$

Furthermore, $$\frac{1}{x^2 + 1} = \frac{1}{x^2 + xy + yz + zx} = \frac{1}{(x + y)(z + x)} \ge \frac{4}{(z + 2x + y)^2}$$

$$\ge \frac{8}{4x^2 + (y + z)^2} = \frac{2}{x^2}\left[1 - \frac{(y + z)^2}{4x^2 + (y + z)^2}\right] = \frac{2}{x^2}\left[1 - \frac{(y + z)^2}{4(x^2 + yz)}\right]$$

$$\implies \sum_{cyc}\frac{1}{x^2 + 1} \ge \sum_{cyc}\frac{2}{x^2}\left[1 - \frac{(y + z)^2}{4(x^2 + yz)}\right]$$

$$\ge \frac{18}{x^2 + y^2 + z^2} - \frac{1}{2}\sum_{cyc}x^2\left[\frac{1}{y^2(y^2 + zx)} + \frac{1}{z^2(z^2 + xy)}\right]$$

And I'm stuck.

Answer 1

We need to prove that:

$$\sum_{cyc}\frac{1}{x^2+1}\geq\frac{2}{3}\left(\sum_{cyc}\frac{x}{\sqrt{x^2+1}}\right)^3$$ or $$\sum_{cyc}xy\sum_{cyc}\frac{1}{(x+y)(x+z)}\geq\frac{2}{3}\left(\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\right)^3.$$ Now, by C-S $$\sum_{cyc}xy\sum_{cyc}\frac{1}{(x+y)(x+z)}=\frac{\sum\limits_{cyc}x(y+z)\sum\limits_{cyc}x}{\prod\limits_{cyc}(x+y)}\geq$$ $$\geq\frac{\left(\sum\limits_{cyc}x\sqrt{y+z}\right)^2}{\prod\limits_{cyc}(x+y)}=\left(\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\right)^2.$$ Id est, it's enough to prove that $$\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}\leq\frac{3}{2},$$ which is true by AM-GM: $$\sum_{cyc}\frac{x}{\sqrt{(x+y)(x+z)}}=\sum_{cyc}\left(\sqrt{\frac{x}{x+y}}\cdot\sqrt{\frac{x}{x+z}}\right)\leq$$ $$\leq\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+y}+\frac{x}{x+z}\right)=\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+y}+\frac{y}{y+x}\right)=\frac{3}{2}.$$ Done!