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.

I think the constant $16$ in the question should be $16/3$. Also, the AM-HM inequality implies
$$\frac{\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}}{4} \ge \frac{4}{3(a+b+c+d)}.$$
