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)
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...

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
847
Back
Top