Determine all possible values a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d)

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In summary, the conversation discusses determining all possible values of the expression S, which is a sum of fractions with arbitrary positive real numbers as the variables. The solution involves using a substitution to simplify the expression and then applying inequalities to find the range of possible values for S. The final conclusion is that the possible values of S lie in the open interval (1,2).
  • #1
lfdahl
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Determine all possible values of

\[S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}\]

when $a,b,c$ and $d$ are arbitrary positive real numbers.
 
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  • #2
lfdahl said:
Determine all possible values of

\[S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d}\]

when $a,b,c$ and $d$ are arbitrary positive real numbers.
[sp]Let $x = a+c$, $y = b+d$. Then $$S = \frac a{a+y} + \frac b{b+x} + \frac c{c+y} + \frac d{d+x}.$$ Taking the first and third terms, $$\begin{aligned} \frac a{a+y} + \frac c{c+y} &= \frac {(a+y)-y}{a+y} + \frac {(c+y)-y}{c+y} \\ &= 2 - \frac y{a+y} - \frac y{c+y} \\ &= 2 - \frac{(a+c)y + 2y^2}{ac + (a+c)y + y^2} \\ &= 2 - \frac{y(x+2y)}{ac + xy + y^2}. \end{aligned} $$ But $0<ac\leqslant \frac14x^2$ (because $a+c=x$ and so the greatest value of $ac$ occurs when $a=c=\frac12x$). Therefore $$2 - \frac{y(x+2y)}{xy+y^2} < \frac a{a+y} + \frac c{c+y} \leqslant 2 - \frac{y(x+2y)}{\frac14x^2 + xy + y^2},$$ $$ 2 - \frac{x+2y}{x+y} < \frac a{a+y} + \frac c{c+y} \leqslant 2 - \frac{y(x+2y)}{\frac14(x+2y)^2},$$ $$ 2 - \frac{(x+y) + y}{x+y} < \frac a{a+y} + \frac c{c+y} \leqslant 2 - \frac{4y}{x+2y},$$ $$ 1 - \frac y{x+y} < \frac a{a+y} + \frac c{c+y} \leqslant \frac{2x}{x+2y}.$$

The exact same procedure applied to the second and fourth terms of $S$ (so that $x$ and $y$ are interchanged) shows that $$ 1 - \frac x{x+y} < \frac b{b+x} + \frac d{d+x} \leqslant \frac{2y}{2x+y}.$$ Add those two sets of inequalities to get $$1 = 2 - \frac x{x+y} - \frac y{x+y} < S \leqslant \frac{2x}{x+2y} + \frac{2y}{2x+y} = \frac{4x^2 + 4xy + 4y^2}{2x^2 + 5xy + 2y^2} < \frac{4x^2 + 10xy + 4y^2}{2x^2 + 5xy + 2y^2} = 2.$$

So $1<S<2$. On the other hand, if $(a,b,c,d) = (1,1,\varepsilon,\varepsilon)$ then \(\displaystyle S = \frac1{2+\varepsilon} + \frac1{2+\varepsilon}+ \frac\varepsilon{1+2\varepsilon}+ \frac\varepsilon{1+2\varepsilon}\), which can be made arbitrarily close to $1$ for small enough $\varepsilon$. If $(a,b,c,d) = (1,\varepsilon,1,\varepsilon)$ then \(\displaystyle S = \frac1{1+2\varepsilon} + \frac\varepsilon{2+\varepsilon}+ \frac1{1+2\varepsilon}+ \frac\varepsilon{2+\varepsilon}\), which can be made arbitrarily close to $2$ for small enough $\varepsilon$.

Finally, the formula for $S$ defines a continuous map from $(\Bbb{R}^+)^4$ to $\Bbb{R}^+$, so its image is connected. Therefore the possible values of $S$ are given by the whole of the open interval $(1,2).$

[/sp]
 
  • #3
Thankyou for an excellent solution, Opalg!(Clapping)
 

Related to Determine all possible values a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d)

1. What is the formula for determining all possible values of a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d)?

The formula for determining all possible values of a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d) is (a+b+c+d)/(a+b+c+d).

2. How many possible values are there for a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d)?

There are an infinite number of possible values for a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d).

3. Can the formula a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d) have negative values?

Yes, the formula a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d) can have negative values depending on the values of a, b, c, and d.

4. Is there a way to simplify the formula a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d)?

Yes, the formula a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d) can be simplified to 1 if a, b, c, and d are all equal.

5. Can the formula a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d) have imaginary solutions?

No, the formula a/(a+b+d)+b/(a+b+c)+c/(b+c+d)+d/(a+c+d) does not have imaginary solutions as it only involves real numbers.

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