Let $a\ge b\ge c\ge d\ge 0$, $c>0$, and $a+b+c+d=1$. I want to show $$ \frac1{\sqrt{2 c}} \le \frac{\sum_{cyc} a [(b+c)(b+d)(c+d)]^{3/2}}{(\sum_{cyc} abc)^{3/2}} \le \frac2{\sqrt{c}}. $$ This is related to this question on Mathoverflow. I can show the expression is at most $4\sqrt{2/3}c^{-1/2}$, but I would like to prove the upper and lower bounds stated above.
In the statement above, $\sum_{cyc}$ means that we sum over all (four) rotations of $a,b,c,d$. E.g. $\sum_{cyc} abc = abc+abd+acd+bcd$.
For the upper bound, assume the expression is maximized at $d=0$, then we just have to show $ \sum_{cyc} (a+b)^{3/2}c^{-1/2} \le 2 c^{-1/2}, $ which follows from $\sum_{cyc} (a+b)^{3/2}\le 2$. I can't actually show that the expression is maximized at $d=0$. The lower bound is sharp when $c=d\to 0$. In particular, if $a=b=1-c$ and $d=c$, then the expression is $\frac{1}{\sqrt{1-2 c}}+\frac{1}{\sqrt{2c}}$, which is cleary at least $\frac{1}{\sqrt{2c}}$.