Is given that $x,y,z$ are positive numbers and $xyz=1$, prove that $$\dfrac{\dfrac{1}{x}}{\sqrt{z^2+1}}+\dfrac{\dfrac{1}{y}}{\sqrt{x^2+1}}+\dfrac{\dfrac{1}{z}} {\sqrt{y^2+1}}>\sqrt{2}.$$ What have I done? First, rewrite our inequality as a function $$f(x,y,z)=\dfrac{1}{x \sqrt{z^2+1}}+\dfrac{1}{y \sqrt{x^2+1}}+\dfrac{1}{z \sqrt{y^2+1}}-\sqrt{2}.$$
Second, I used this: $a^2+b^2 \geq 2c.$ So, we got $$\dfrac{1}{x \sqrt{z^2+1}}+\dfrac{1}{y \sqrt{x^2+1}}+\dfrac{1}{z \sqrt{y^2+1}}-\sqrt{2} \geq \dfrac{1}{x \sqrt{2z}}+\dfrac{1}{y\sqrt{(2x}}+\dfrac{1}{z*\sqrt{2y}}-\sqrt{2}.$$
Further I used this: $\dfrac{1}{\sqrt{ab}} \geq \dfrac{2}{a+b}.$ We got $$\dfrac{1}{x\sqrt{2z}}+\dfrac{1}{y\sqrt{2x}}+\dfrac{1}{z\sqrt{2y}}-\sqrt{2} \geq \dfrac{2}{x(2+z)}+\dfrac{2}{y(2+x)}+\dfrac{2}{z(2+y)}-\sqrt{2}.$$
The next step is to use this: $$\dfrac{a}{b}+\dfrac{c}{d}+\dfrac{e}{f} \geq 3 \sqrt[3]{\dfrac{ace}{dbf}}.$$ And we got $$\dfrac{2}{x(2+z)}+\dfrac{2}{y(2+x)}+\dfrac{2}{z(2+y)}-\sqrt{2} \geq 3\sqrt[3]{\dfrac{8}{xyz(2+x)(2+y)(2+z)}}-\sqrt{2} .$$
What I did next is raised both sides of the inequality to the sixth power $$3^6 \dfrac{8}{xyz(2+z)(2+x)(2+y)} \geq 2 \sqrt{2}$$ and $$3^6 \dfrac{8}{xyz(2+z)(2+x)(2+y)}-2 \sqrt{2} \geq 0.$$ I don't know what to do next and how to show that it is greater than zero? Any hint would help a lot! Thanks!
If I understood right your problem.
Let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{c}{a}$ and by AM-GM we obtain: $$\sum_{cyc}\frac{1}{x\sqrt{z^2+1}}=\sum_{cyc}\frac{1}{\frac{a}{b}\sqrt{1+\frac{c^2}{a^2}}}=\sum_{cyc}\frac{b}{\sqrt{a^2+c^2}}=$$ $$=\sum_{cyc}\frac{2b^2}{2\sqrt{b^2(a^2+c^2)}}\geq\sum_{cyc}\frac{2b^2}{b^2+a^2+c^2}=2>\sqrt2.$$ Because $$\sum_{cyc}\frac{2b^2}{b^2+a^2+c^2}=\frac{2b^2}{a^2+b^2+c^2}+\frac{2c^2}{a^2+b^2+c^2}+\frac{2a^2}{a^2+b^2+c^2}=2.$$ Your solution is wrong because by AM-GM $$\frac{1}{x\sqrt{1+z^2}}\leq\frac{1}{x\cdot\sqrt{2z}}.$$