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

  • Thread starter Thread starter lfdahl
  • Start date Start date
AI Thread Summary
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 \( (1, 2) \).
lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
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.
 
Mathematics news on Phys.org
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)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.

Similar threads

Replies
10
Views
1K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Replies
2
Views
1K
Replies
1
Views
861
Back
Top