I encountered the following problem and was hinted to consider the edge case.

Determine all possible values of $$S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}$$ where $a$, $b$, $c$ and $d$ are arbitrary positive numbers.

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    $\begingroup$ Little hint: the function is continuous on the connected set $(0,+\infty)^4$, so the image is an interval. $\endgroup$ – Julien Mar 3 '13 at 18:35

The expression is homogeneous of order zero, meaning it does not change if the vector $(a,b,c,d)$ is multiplied by a constant. So you might as well assume $a+b+c+d=1$, and we are looking at the interior points of the simplex with corners at $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$ and $(0,0,0,1)$. Further, note we can now write $$ S=\frac{a}{1-c}+\frac{b}{1-d}+\frac{c}{1-a}+\frac{d}{1-b}.$$ If you fix $b$ and $d$ then $a+c$ is fixed, and it is easy to see that the first and third terms are each concave functions of $a$ (or, equivalently, of $c$). Hence so is their sum, so the minimum appears at the ends of the allowable interval, i.e., where either $a$ or $c$ vanishes. By symmetry, we may look at just the case $c=0$. A similar argument applies the the second and fourth terms and the variables $b$ and $d$, so we may set $d=0$. But with $c=d=0$ the whole expression reduces to $a+b=1$, so $S$ has the minimum value $1$ on the closed simplex. Of course, this will not be achieved in the interior of the simplex, but the infimum is still $1$.

Edit: Three years have passed, and now a request has surfaced to finish this. So here goes:

It remains to find the supremum. If we fix $b$ and $d$, as noted the sum of the first and third terms $a/(1-c)+c/(1-a)$ is a concave function on the interval $a+c=1-b-d$, $a>0$, $c>0$. Moreover, this function is symmetric in $a$, $c$, so it has its maximum at the midpoint of the interval, where $a=c$. The same goes for the sum of the second and fourth terms, in the variables $b$, $d$. In other words, for any point in the simplex, there is another point with $a=c$ and $b=d$ with a greater value for the sum $S$. For such a point, we get $$ \frac S2=\frac a{1-a}+\frac b{1-b},$$ and now we have the constraint $a+b=\frac12$, $a>0$, $b>0$. But this function is convex, so its maximum at the endpoints of the interval. Putting $a=0$, $b=\frac12$ or $a=\frac12$, $b=0$, we get $S=2$. Of course, these values aren't achieved, but still, the supremum is $2$.

  • $\begingroup$ One too many 0's in vertex $(1,0,0,0,0)$. But a good suggestion. +1 $\endgroup$ – coffeemath Mar 3 '13 at 18:48
  • $\begingroup$ Thank you, Zev, for the posting advice. And thank you, Harald, for the solution. I expressed S in the way Harald does, but failed to see the convexity in each fraction. Thanks, Harald. $\endgroup$ – Hans Mar 3 '13 at 19:07
  • $\begingroup$ As I was too hasty in posting the answer, you should not be too hasty in thanking me. Better put the thinking cap back on. $\endgroup$ – Harald Hanche-Olsen Mar 3 '13 at 19:17
  • $\begingroup$ I am also leaving the wrong answer up because it has already attracted some attention, and having it disappear without an explanation will just confuse everybody. $\endgroup$ – Harald Hanche-Olsen Mar 3 '13 at 19:36
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    $\begingroup$ The maximum is not $4/3$, since if $a=c=1$ while $b=d=1/2$ we have $S=7/5=1.4>4/3=1.33..$ $\endgroup$ – coffeemath Mar 4 '13 at 5:45

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