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

Question

Prove or disprove that the inequality $$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$ is valid if $x,y,z$ are positive numbers and $$xyz=1.$$

My solution is: Let $$x=\dfrac{a}{b}, y=\dfrac{b}{c}, z=\dfrac{c}{a}.$$ So we have $$\dfrac{1}{\sqrt{1+\dfrac{a}{b}}}+\dfrac{1}{\sqrt{1+\dfrac{b}{c}}}+\dfrac{1}{\sqrt{1+\dfrac{c}{a}}}=\dfrac{1}{\sqrt{\dfrac{a+b}{b}}}+\dfrac{1}{\sqrt{\dfrac{b+c}{c}}}+\dfrac{1}{\sqrt{\dfrac{c+a}{a}}}=\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}=\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}}.$$ Then we can use this: $$\dfrac{1}{\sqrt{xy}}\geq\dfrac{2}{x+y}.$$ So we have $$\sqrt{\dfrac{bb}{b(a+b)}}+\sqrt{\dfrac{cc}{c(c+b)}}+\sqrt{\dfrac{aa}{a(a+c)}} \geq \dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c}.$$ Then we can use this: $$\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z} \geq 3 \sqrt[3]{\dfrac{abc}{xyz}}.$$ So we have $$\dfrac{2b}{2b+a}+\dfrac{2c}{2c+b}+\dfrac{2a}{2a+c} \geq 3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}}.$$ The question is what should I do next? Is it already obvious that $$3 \sqrt[3] {\dfrac{8abc}{(2b+a)(2c+b)(2a+c)}} \geq 1?$$ Any hint would help a lot! Thanks!

Answer 1

Let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.

Thus, $z=\frac{c}{a}$ and $$\sum_{cyc}\frac{1}{\sqrt{1+x}}=\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a+b}}\geq\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$

Answer 2

This is almost the same as the other answer and just slightly different: $\sqrt{\dfrac{b}{a+b}}+\sqrt{\dfrac{c}{c+b}}+\sqrt{\dfrac{a}{a+c}}\ge\sqrt{\dfrac{b}{a+b+c}}+\sqrt{\dfrac{c}{a+b+c}}+\sqrt{\dfrac{a}{a+b+c}}=\dfrac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a+b+c}}> 1.$ Done.