Question 1157229
<pre>

Here is the best I could come up with (see CAVEAT near end):

{{{ f = ((1/a)+(1/b)+(1/c)) / ((1/(a+b))+(1/(a+c))+(1/(b+c))) }}}

For a=b=c:

{{{ f = ((3/a))/((3/2a)) = (3/a)*(2a/3) = 2 }}}

So when the numbers are all equal, we get f(a,b,c) = 2.  I aim to show that 2 is the minimum value.  

Let a be the smallest number (due to symmetry, no loss of generality).

There are two cases to explore, the first case we let one of the other numbers be bigger than a:  

  {{{ a=a }}}
  {{{ b=a+epsilon }}}
  {{{ c=a }}}

(f=N/D)
N = {{{ (1/a)+(1/b)+(1/c) = (1/a)+(1/(a+epsilon))+1/a = (3a+2*epsilon)/(a*(a+epsilon)) }}}

D = {{{ (1/(a+b))+(1/(a+c))+(1/(b+c)) = (1/(2a+epsilon))+(1/(2a))+(1/(2a+epsilon )) }}}
  = {{{ (6a+epsilon)/(2a*(2a+epsilon)) }}} 

N/D = N*(1/D) = {{{ ((3a+2*epsilon)*(2a*(2a+epsilon))) / ( (a*(a+epsilon))*(6a+epsilon)) }}}

= {{{ (12a^2+14a*epsilon + 4*epsilon^2) / (6a^2+7a*epsilon+epsilon^2) }}} 

For any {{{epsilon > 0}}} this last expression is > 2.   EDIT 4/28: to be precise, it is {{{ 2 + 2*epsilon^2/( (6a^2+7a*epsilon+epsilon^2) ) }}} 


The 2nd case is to let both b and c be larger than a.  I am keeping it simple and letting b=c.  This is equivalent to writing {{{ a=b-epsilon}}}, {{{b=b}}}, {{{c=b}}}:

The resulting N*(1/D) expression is
{{{ N*(1/D) =  (12b^2-14b*epsilon+4*epsilon^2) / (6b^2-7b*epsilon+epsilon^2) }}} and this expression is > 2 as well.  EDIT 4/28: the precise value is {{{ 2 + 2*epsilon^2/( (6b^2-7b*epsilon+epsilon^2) ) }}} 


Therefore  {{{ highlight( f_min(a,b,c) = 2 ) }}} 

CAVEAT:
This proves f_min=2 holds for three of four scenarios: (1) {{{a=b=c}}}, (2) {{{a<b}}}, {{{c=a}}}, and (3)  {{{a<b}}}, {{{b=c}}}.   My proof does NOT cover the case {{{a<b<c }}}, although I'm certain it will hold true.  I leave this case to the student (or motivated tutor).   To handle this case, let {{{ a=a}}}, {{{ b=a+epsilon}}}, {{{ c = a+delta }}}.  The math will look a lot messier... 

Perhaps another tutor will find a simpler solution(?) 


EDIT 4/28:  The proof for {{{a<b<c }}} is very challenging (at least for me).  It has some subtlties that I'm unable to see.  Although not a proof, a SPECIFIC EXAMPLE is easy to illustrate:

Let a=a
    b=2a 
    c=4a

{{{ f = ((1/a)+(1/b)+(1/c)) / ((1/(a+b))+(1/(a+c))+(1/(b+c))) }}}
= {{{ ((1/a)+ (1/(2a))+ (1/(4a))) / ((1/(a+2a))+(1/(a+4a))+(1/(2a+4a))) }}}
= {{{  (7/(4a)) / ((1/(3a))+(1/(5a))+(1/(6a))) }}}
= {{{  ((7/(4a)))/((21/(30a))) }}}
= {{{  (7/4) * (30/21) }}}
= {{{   2.5 }}} > 2