Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$. 
If $a,b,c,d > 0$ and distinct then show that
  $$
\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c}
\ge \frac{16}{3(a+b+c+d)}
$$

I tried using HM < AM inequality but am missing on $16$. Probably I am mistaken in solving.

This fails for $a=1,$ $b=2,$ $c=3,$ & $d=4$.  The left-hand side becomes $\frac{275}{504}=0.5456349...$ and the right-hand side is $\frac{16}{10}=1.6$
