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.

Hi I don't think this is true. For example if $a=b=c=d=1$ then we have $4/3 >= 4$ which is false.
