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.

You can assume $a+b+c+d=1$, so we need to prove $$E:=\sum {1\over 1-x} \geq 16/3$$
Now, since for $x\in (0,1)$ we have $${1\over 1-x}\geq {16x\over 9}+{8\over 9}$$ we get $$E\geq {16\over 9}(a+b+c+d)+4{8\over 9} = {48\over 9}$$
