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.

What? So the left hand side is the AM of $1/(b+c+d), 1/(c+d+a), 1/(d+a+b)$ and $1/(a+b+c)$ i.e. $(1/(b+c+d)+1/(a+c+d)+1/(a+b+d)+1/(a+b+c))/4$. On the RHS we have $4$ over the sum of the reciprocals of these terms i.e.   $4/((b+c+d)+(c+d+a)+(d+a+b)+(a+b+c))$ which does indeed come out as $4/3(a+b+c+d)$ Which then (multiplying both sides by 4) yields the result $1/(b+c+d)+1/(a+c+d)+1/(a+b+d)+1\(a+b+c) \geq 16 / 3(a+b+c+d)$. Isn't that cool?
