Selection Examination

From the inequality $2yz\le y^2+z^2$ we have that $x^2+2yz+2\le x^2+y^2+z^2+2$, so $$\frac{x}{\sqrt{x^2+2yz+2}}\ge \frac{x}{\sqrt{x^2+y^2+z^2+2}}$$ Working similarly and adding we have that $$\frac{x}{\sqrt{x^2+2yz+2}}+\frac{y}{\sqrt{y^2+2zx+2}}+\frac{z}{\sqrt{z^2+2xy+2}}\ge \frac{x+y+z}{\sqrt{x^2+y^2+z^2+2}}$$ Therefore, it suffices to prove that $$\frac{x+y+z}{\sqrt{x^2+y^2+z^2+2}}\ge 1\iff (x+y+z{)}^2\ge x^2+y^2+z^2+2\iff xy+yz+zx\ge 1$$ However, from the given condition we have $\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=9$, and from the Cauchy-Schwarz inequality we have $$(xy+yz+zx)\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\ge 9,\text{ so }xy+yz+zx\ge 1,\text{ which is the desired result.}$$

2ºς τρόπος: Using Holder's inequality we have: $$\left(\sum _{cyc}\frac{x}{\sqrt{x^2+2yz+2}}\right)\left(\sum _{cyc}\frac{x}{\sqrt{x^2+2yz+2}}\right)\left(\sum _{cyc}x(x^2+2yz+2)\right)\ge (x+y+z{)}^3$$ Therefore, it suffices to prove that $$\begin{array}{rl}\frac{(x+y+z{)}^3}{{\sum }_{cyc}x(x^2+2yz+2)}& \ge 1\\ & \iff (x+y+z{)}^3\ge x^3+y^3+z^3+6xyz+2(x+y+z)\\ & \iff x^3+y^3+z^3+3(x+y)(y+z)(z+x)\ge x^3+y^3+z^3+6xyz+2(9xyz)\\ & \iff (x+y)(y+z)(z+x)\ge 8xyz\end{array}$$ but the last one holds since $$x+y\ge 2\sqrt{xy},y+z\ge 2\sqrt{yz},z+x\ge 2\sqrt{zx}.$$