Proof : By Cauchy Schwarz Inequality $\Rightarrow \sum_{cyc} \frac{1}{\sqrt{x+2y+6}}\leq 3\sum_{cyc} \frac{1}{x+2y+6}$ (1) and $\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{y+2}\geq\frac{9}{x+2y+6}$ $\sum_{cyc} \Big(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{y+2}\Big)\geq \sum_{cyc}\frac{9}{x+2y+6}$ $\Rightarrow\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\geq 3\sum_{cyc}\frac{1}{x+2y+6}$ (2) Consider $\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}$=$\frac{xy+yz+zx+4(x+y+z)+12}{2(xy+yz+zx)+4(x+y+z)+9}$ By A.M.-G.M. $\Rightarrow xy+yz+zx\geq 3$ ($xyz=1$) $\Rightarrow \frac{xy+yz+zx+4(x+y+z)+12}{2(xy+yz+zx)+4(x+y+z)+9}\leq 1$ $\Rightarrow \frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}\leq 1$ (3) Combine (1),(2) and (3) So, $\sum_{cyc} \frac{1}{\sqrt{x+2y+6}}\leq 1 \blacksquare$

Proof : $\frac{x}{\sqrt{x^2+4\sqrt{y}+4\sqrt{z}}}\geq \frac{x}{x+y+z}$ $\iff \sqrt{x^2+4\sqrt{y}+4\sqrt{z}}\leq x+y+z $ $\iff x^2+4\sqrt{y}+4\sqrt{z}\leq (x+y+z)^2 $ $\iff 4\sqrt{y}+4\sqrt{z}\leq (y+z)^2+2xy+2xz $ $\iff 4(\sqrt{y}+\sqrt{z})\leq (y+z)^2+2(\frac{1}{y}+\frac{1}{z}) $ (1) We use Cauchy Schwarz Inequality implies $y+z\geq\frac{(\sqrt{y}+\sqrt{z})^2}{2}\\ \Rightarrow (y+z)^2\geq\frac{(\sqrt{y}+\sqrt{z})^4}{4}$ (2) We use A.M.-G.M. Inequality implies $\frac{1}{y}+\frac{1}{z}\geq\frac{2}{\sqrt{yz}}$ and $(\sqrt{y}+\sqrt{z})^2\geq4\sqrt{yz}$ So, $\frac{1}{y}+\frac{1}{z}\geq \frac{8}{(\sqrt{y}+\sqrt{z})^2}$ (3) Combine (2),(3): $(y+z)^2+2(\frac{1}{y}+\frac{1}{z})\geq\frac{16}{(\sqrt{y}+\sqrt{z})^2}+\frac{(\sqrt{y}+\sqrt{z})^4}{4}\\ \geq4(\sqrt{y}+\sqrt{z})$ (By A.M.-G.M.) Inequality (1) is true. So, $\frac{x}{\sqrt{x^2+4\sqrt{y}+4\sqrt{z}}}\geq \frac{x}{x+y+z}$ Similarly, $\frac{y}{\sqrt{y^2+4\sqrt{z}+4\sqrt{x}}}\geq \frac{y}{x+y+z}$ and $\frac{z}{\sqrt{z^2+4\sqrt{x}+4\sqrt{y}}}\geq \frac{z}{x+y+z}$ $\Rightarrow \sum_{cyc} \frac{x}{\sqrt{x^2+4\sqrt{y}+4\sqrt{z}}}\geq 1 \blacksquare$