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

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The expression \( S = \frac{a}{a+b+d}+\frac{b}{a+b+c}+\frac{c}{b+c+d}+\frac{d}{a+c+d} \) is analyzed for arbitrary positive real numbers \( a, b, c, \) and \( d \). Through manipulation and inequalities, it is established that \( 1 < S < 2 \). Specific cases demonstrate that \( S \) can approach values arbitrarily close to 1 and 2. The continuity of the function implies that all values in the open interval \( (1, 2) \) are achievable. Thus, the possible values of \( S \) are the entire open interval \( (1, 2) \).
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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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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 $$S = \frac1{2+\varepsilon} + \frac1{2+\varepsilon}+ \frac\varepsilon{1+2\varepsilon}+ \frac\varepsilon{1+2\varepsilon}$$, which can be made arbitrarily close to $1$ for small enough $\varepsilon$. If $(a,b,c,d) = (1,\varepsilon,1,\varepsilon)$ then $$S = \frac1{1+2\varepsilon} + \frac\varepsilon{2+\varepsilon}+ \frac1{1+2\varepsilon}+ \frac\varepsilon{2+\varepsilon}$$, 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]
 
Thankyou for an excellent solution, Opalg!(Clapping)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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