# Prove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1.$

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!

• Here is a MathJax tutorial Commented Dec 5, 2022 at 21:03
• J. W. Tanner thank you very much! Commented Dec 5, 2022 at 21:06
• I did the mathjax but I don't know what are last notations that you wrote @AliceMalinova
– PNT
Commented Dec 5, 2022 at 21:43
• PNT thank you very much! it's my mistake, there is \sqrt instead of \dfrac Commented Dec 5, 2022 at 21:47
• There is an error early in your calculations. $z^2+1 \ge 2 z$ implies that $\frac{1}{x\sqrt{z^2+1}} \color{red}{\le} \frac{1}{x\sqrt{2z}}$. Commented Dec 5, 2022 at 22:02

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 $$\frac{1}{x\sqrt{1+z^2}}\leq\frac{1}{x\cdot\sqrt{2z}}.$$
• @Alice Malinova It's just homogenization because $\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=1$. I added something. See now. Commented Dec 5, 2022 at 21:42