Show $\frac{1}{b+c+d} + \frac{1}{a+c+d} + \frac{1}{a+b+d} + \frac{1}{a+b+c} \ge \frac{16}{3(a+b+c+d)}$.

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $\frac{16}{10}=1.6$

Answer 4

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 $$E\geq {16\over 9}(a+b+c+d)+4{8\over 9} = {48\over 9}$$

Answer 5

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?