Exploring an inequality between $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} $ and $\frac{3}{1+(abc)^{1/3}}$ if $a,b, c>0$ An AM-HM inequality for three positive numbers leads to 
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge \frac{3}{{1+\frac{a+b+c}{3}}}. ~~~~(1)$$ 
Next, the well known AM-GM inequality
$$\frac{a+b+c}{3} \ge (abc)^{1/3} ~~~~(2)$$  becomes incompatible with (1) to
claim any inequality between $$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} ~ \mbox{and} ~ \frac{3}{1+(abc)^{1/3}}.$$
Hence, the question is: Can one establish an inequality between these two expressions?
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}~ \mbox{and}~ \frac{3}{1+(abc)^{1/3}} ~\mbox{if}~ a,b, c>0$$.
 A: If $f''(x)>0$ in a domain $D$, the Jensen's inequality claims  that the mean of functions is greater or equal to the function of mean:
$$\frac{f(x)+f(y)+f(z))}{3} \ge f\left(\frac{x+y+z}{3} \right) , \forall ~x,y,z \in D.~~~~(1)$$ The sign of the inequality reverses if $f''(x) < 0, \forall x \in D$.
Consider $f(x)=\frac{1}{1+e^{x}}$, then $f''(x)=e^{x}\frac{e^{x}-1}{(e^x+1)^2}>0,~ \mbox{if}~ x>0;$
then for three positive numbers $x,y,z$ from (1), we can write
$$\frac{1}{1+e^x}+ \frac{1}{1+e^y}+\frac{1}{1+e^z} \ge \frac{3}{1+e^{(x+y+z)/3}}.~~~~(2)$$
Letting $e^x, e^y, e^z$ equal to $a,b,c \ge 1$, respectively, we establish an interesting inequality from (1), that
$$\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} \ge\frac{3}{1+(abc)^{1/3}},~~~ \mbox{if}~ a,b,c \ge 1. ~~~~(3).$$ Next, when $0 < a,b,c \le 1$, the sign of the inequality reverses. For other combinations of $a,b,c>0$; the inequality (3) may or may not hold. Equality holds when $a=b=c=1.$ We have not included $a,b,c=0$ because these are attained at $x,y,z=-\infty$, however, the equality hold also at $a=b=c=0.$ 
