Earlier
the expressions $$\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[n]{1+b}}+\frac{1}{\sqrt[n]{1+c}}$$ and $$\frac{3}{\sqrt[n]{1+(abc)^{1/3}}}$$ have been found to display two inequalities for $n=1$ when $a,b,c \in [0,1]$ or in $[1,\infty)$. Now the question is if there exist more general inequalities between them if $n>0$ ?
Take $f(x)=\frac{1}{(1+e^x)^{1/n}}, n>0$. then $f''(x)=\frac{e^x(e^x -n)}{n^2(1+e^x)^{1/n+2}}>0,~ \mbox{if}~ e^x>n.$ Then Jensen's inequality $$\frac{f{x}+f(y)+f(z)}{3} \ge f\left(\frac{x+y+z)}{3}\right)~~~~(1)$$ holds. So we have $$\frac{1}{\sqrt[n]{1+e^x}}+\frac{1}{\sqrt[n]{1+e^y}}+ \frac{1}{\sqrt[n]{1+e^z}}\ge \frac{1}{\sqrt[n]{1+e^{(x+y+z)/3}}},~\mbox{if}~ e^x,e^y,e^z >n, n>0~~~~(2).$$ Hence, we can re-write it as $$\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[n]{1+b}}+\frac{1}{\sqrt[n]{1+c}} \ge \frac{3}{\sqrt[n]{1+(abc)^{1/3}}},~ \mbox{if}~ a,b,c>n,n>0.~~~~(3).$$ The sign of inequality(3) reverses if $0<a,b,c <n$. The equality holds when $a=b=c=n$.
