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

lfdahl

Well-known member
Nov 26, 2013
719
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.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,684
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.
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 $(1,2).$

 

lfdahl

Well-known member
Nov 26, 2013
719
Thankyou for an excellent solution, Opalg !(Clapping)