Prove that the inequality $$\dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}} \geq 1$$ is true if $x, y, z>0$ and
$1)$ $xyz=1$
$2)$ $xyz=8^3.$
For the first case it turns out to be very easy to get the desired inequality with this substitution $$x=\dfrac{a}{b}, \ y=\dfrac{b}{c}, \ z=\dfrac{c}{a}, \ \ a,b,c >0.$$ So we have $$ \dfrac{1}{\sqrt{x+1}}+\dfrac{1}{\sqrt{y+1}}+\dfrac{1}{\sqrt{z+1}} = \dfrac{1}{\sqrt{\dfrac{a}{b}+1}}+\dfrac{1}{\sqrt{\dfrac{b}{c}+1}}+\dfrac{1}{\sqrt{\dfrac{c}{a}+1}}\\ = \dfrac{1}{\sqrt{\dfrac{a+b}{b}}}+\dfrac{1}{\sqrt{\dfrac{b+c}{c}}}+\dfrac{1}{\sqrt{\dfrac{c+a}{a}}}\\ = \dfrac{\sqrt{b}}{\sqrt{a+b}}+\dfrac{\sqrt{c}}{\sqrt{b+c}}+\dfrac{\sqrt{a}}{\sqrt{a+c}}.$$ Let us use this inequality $$ \sqrt{a}+\sqrt{b}=\sqrt{a+b+2\sqrt{ab}} \geq \sqrt{a+b}.$$ So we have $$\dfrac{\sqrt{b}}{\sqrt{a+b}}+\dfrac{\sqrt{c}}{\sqrt{b+c}}+\dfrac{\sqrt{a}}{\sqrt{a+c}} \geq \dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}+\dfrac{\sqrt{c}}{\sqrt{b}+\sqrt{c}+\sqrt{a}}+\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{c}+\sqrt{b}} = \dfrac{\sqrt{b}+\sqrt{c}+\sqrt{a}}{\sqrt{a}+\sqrt{c}+\sqrt{b}} = 1.$$ But for the second case, when $xyz=8^3$, we cannot use the same substitution as we will get $1$ multiplying all the variables. Any idea? Thanks in advance!
A proof for $xyz=512.$
Let $x=\frac{8bc}{a^2}$ and $y=\frac{8ac}{b^2},$ where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{8ab}{c^2}$ and by Holder and AM-GM we obtain: $$\sum_{cyc}\tfrac{1}{\sqrt{x+1}}=\sum_{cyc}\tfrac{a}{\sqrt{8bc+a^2}}=\sqrt{\tfrac{\left(\sum\limits_{cyc}\tfrac{a}{\sqrt{8bc+a^2}}\right)^2\sum\limits_{cyc}a(8bc+a^2)}{\sum\limits_{cyc}a(8bc+a^2)}}\geq\sqrt{\tfrac{(a+b+c)^3}{\sum\limits_{cyc}a(8bc+a^2)}}\geq1.$$