For positive real numbers $a,b,c,d$, with $abcd = 1$, determine all values taken by the expression \[\frac {1+a+ab} {1+a+ab+abc} + \frac {1+b+bc} {1+b+bc+bcd} +\frac {1+c+cd} {1+c+cd+cda} +\frac {1+d+da} {1+d+da+dab}.\] (Dan Schwarz)
Problem
Source: Stars of Mathematics 2011 - Juniors - Problem 1
Tags: algebra proposed, algebra
Farenhajt
11.10.2012 00:42
Expand the first fraction by $d$, the second by $ad$, the third by $abd$ and simplify to get $E=3$.
mathreyes
18.11.2012 20:27
You can obtain this result even if you set $x_{1}x_{2}\cdots x_{n}=1$ and write $n$ fractions geerated by generalization of above.
chxris
19.11.2012 15:12
See that, as $abcd=1$, \[ \frac{1+a+ab}{1+a+ab+abc}+\frac{1+b+bc}{1+b+bc+bcd}+\frac{1+c+cd}{1+c+cd+cda}+\frac{1+d+da}{1+d+da+dab} \] is equivalent to \[\frac{1+a+ab}{abc(1+d+da+dab)}+\frac{1+b+bc}{bc(1+d+da+dab)}+\frac{1+c+cd}{c(1+d+da+dab)}+\frac{1+d+da}{1+d+da+dab}\] $= \frac{1}{abc(1+d+da+dab)}((1+a+ab)+a(1+b+bc)+ab(1+c+cd)+abc(a+d+da))$ $=\frac{1}{1+a+ab+abc}(3+3a+3ab+3abc) =3$ Hence, the expression above is equal to 3, which means that it is the only value the expression can take.
ionbursuc
03.03.2013 21:32
Generalization Let ${{\alpha }_{1}},{{\alpha }_{2}},...{{\alpha }_{n+1}},{{a}_{1}},{{a}_{2}},...,{{a}_{n+1}}\in \left( 0,\infty \right)$ with ${{a}_{1}}{{a}_{2}}...{{a}_{n+1}}=1$ and $k\in \left\{ 1,2,...,n \right\}$.Prove that $\sum\limits_{cyc}{\frac{{{\alpha }_{k}}{{a}_{k}}{{a}_{k+1}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}=n+2-k}.$
Proof
Exist ${{x}_{1}},{{x}_{2}},...,{{x}_{n+1}}$ so ${{a}_{1}}=\frac{{{x}_{1}}}{{{x}_{2}}},{{a}_{2}}=\frac{{{x}_{2}}}{{{x}_{3}}},...,{{a}_{n}}=\frac{{{x}_{n}}}{{{x}_{n+1}}},{{a}_{n+1}}=\frac{{{x}_{n+1}}}{{{x}_{1}}}$
$\sum\limits_{cyc}{\frac{{{\alpha }_{k}}{{a}_{k}}{{a}_{k+1}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}=}$
$\sum\limits_{cyc}{\frac{{{\alpha }_{k}}\frac{{{x}_{k}}}{{{x}_{n+1}}}+...+{{\alpha }_{n}}\frac{{{x}_{n}}}{{{x}_{n+1}}}+{{\alpha }_{n+1}}}{{{\alpha }_{1}}\frac{{{x}_{1}}}{{{x}_{n+1}}}+{{\alpha }_{2}}\frac{{{x}_{2}}}{{{x}_{n+1}}}+...+{{\alpha }_{n}}\frac{{{x}_{n}}}{{{x}_{n+1}}}+{{\alpha }_{n+1}}}=}\sum\limits_{cyc}{\frac{{{\alpha }_{k}}{{x}_{k}}+...+{{\alpha }_{n}}{{x}_{n}}+{{\alpha }_{n+1}}{{x}_{n+1}}}{{{\alpha }_{1}}{{x}_{1}}+{{\alpha }_{2}}{{x}_{2}}+...+{{\alpha }_{n}}{{x}_{n}}+{{\alpha }_{n+1}}{{x}_{n+1}}}=}n+2-k.$
Particular cases
1.For $n=2,k=2$ we obtain:
Let ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},{{a}_{1}},{{a}_{2}},{{a}_{3}}\in \left( 0,\infty \right)$ with ${{a}_{1}}{{a}_{2}}{{a}_{3}}=1$ .Prove that
$\frac{{{\alpha }_{2}}{{a}_{2}}+{{\alpha }_{3}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}+{{\alpha }_{2}}{{a}_{2}}+{{\alpha }_{3}}}+\frac{{{\alpha }_{3}}{{a}_{3}}+{{\alpha }_{1}}}{{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}+{{\alpha }_{3}}{{a}_{3}}+{{\alpha }_{1}}}+\frac{{{\alpha }_{1}}{{a}_{1}}+{{\alpha }_{2}}}{{{\alpha }_{3}}{{a}_{3}}{{a}_{1}}+{{\alpha }_{1}}{{a}_{1}}+{{\alpha }_{2}}}=2$.
2.For $n=3,k=2$ we obtain:
Let ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},{{\alpha }_{4}},{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}\in \left( 0,\infty \right)$ with ${{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}=1$ .Prove that
$\frac{{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}+{{\alpha }_{3}}{{a}_{3}}+{{\alpha }_{4}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}{{a}_{3}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}+{{\alpha }_{3}}{{a}_{3}}+{{\alpha }_{4}}}+\frac{{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}+{{\alpha }_{4}}{{a}_{4}}+{{\alpha }_{1}}}{{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}{{a}_{4}}+{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}+{{\alpha }_{4}}{{a}_{4}}+{{\alpha }_{1}}}+$
$\frac{{{\alpha }_{4}}{{a}_{4}}{{a}_{1}}+{{\alpha }_{1}}{{a}_{1}}+{{\alpha }_{2}}}{{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}{{a}_{1}}+{{\alpha }_{4}}{{a}_{4}}{{a}_{1}}+{{\alpha }_{1}}{{a}_{1}}+{{\alpha }_{2}}}+\frac{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}+{{\alpha }_{2}}{{a}_{2}}+{{\alpha }_{3}}}{{{\alpha }_{4}}{{a}_{4}}{{a}_{1}}{{a}_{2}}+{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}+{{\alpha }_{2}}{{a}_{2}}+{{\alpha }_{3}}}=3.$
3.For $n=3,k=3$ we obtain:
Let ${{\alpha }_{1}},{{\alpha }_{2}},{{\alpha }_{3}},{{\alpha }_{4}},{{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}\in \left( 0,\infty \right)$ with ${{a}_{1}}{{a}_{2}}{{a}_{3}}{{a}_{4}}=1$ .Prove that
$\frac{{{\alpha }_{3}}{{a}_{3}}+{{\alpha }_{4}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}{{a}_{3}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}+{{\alpha }_{3}}{{a}_{3}}+{{\alpha }_{4}}}+\frac{{{\alpha }_{4}}{{a}_{4}}+{{\alpha }_{1}}}{{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}{{a}_{4}}+{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}+{{\alpha }_{4}}{{a}_{4}}+{{\alpha }_{1}}}+$
$\frac{{{\alpha }_{1}}{{a}_{1}}+{{\alpha }_{2}}}{{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}{{a}_{1}}+{{\alpha }_{4}}{{a}_{4}}{{a}_{1}}+{{\alpha }_{1}}{{a}_{1}}+{{\alpha }_{2}}}+\frac{{{\alpha }_{2}}{{a}_{2}}+{{\alpha }_{3}}}{{{\alpha }_{4}}{{a}_{4}}{{a}_{1}}{{a}_{2}}+{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}+{{\alpha }_{2}}{{a}_{2}}+{{\alpha }_{3}}}=2.$
4.For $n=3,k=2$ we obtain:
Let ${{\alpha }_{1}}={{\alpha }_{2}}={{\alpha }_{3}}={{\alpha }_{4}}=1,{{a}_{1}}=a,{{a}_{2}}=b,{{a}_{3}}=c,{{a}_{4}}=d,a,b,c,d\in \left( 0,\infty \right)$ with $abcd=1$.
Calculate \[\frac {1+a+ab} {1+a+ab+abc} + \frac {1+b+bc} {1+b+bc+bcd} +\frac {1+c+cd} {1+c+cd+cda} +\frac {1+d+da} {1+d+da+dab}.\]
(Dan Schwarz,Stars of Mathematics 2011 - Juniors - Problem 1)
ionbursuc
04.03.2013 18:15
Generalization 2 Let${{\alpha }_{1}},{{\alpha }_{2}},...{{\alpha }_{n+1}},{{a}_{1}},{{a}_{2}},...,{{a}_{n+1}}\in \left( 0,\infty \right) $with ${{a}_{1}}{{a}_{2}}...{{a}_{n+1}}=1$ and ${{p}_{1}},{{p}_{2}},...,{{p}_{n+1}}\in \mathbb{R}.$ Prove that $\sum\limits_{cyc}{\frac{{{p}_{1}}{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{p}_{2}}{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{p}_{n}}{{\alpha }_{n}}{{a}_{n}}+{{p}_{n+1}}{{\alpha }_{n+1}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}=}$${{{p}_{1}}}+{{p}_{2}}+...+{{p}_{n+1}}.$ (${{p}_{1}},{{p}_{2}},...,{{p}_{n+1}}$ are fixed )
Proof
Exist${{x}_{1}},{{x}_{2}},...,{{x}_{n+1}}>0$ so ${{a}_{1}}=\frac{{{x}_{1}}}{{{x}_{2}}},{{a}_{2}}=\frac{{{x}_{2}}}{{{x}_{3}}},...,{{a}_{n}}=\frac{{{x}_{n}}}{{{x}_{n+1}}},{{a}_{n+1}}=\frac{{{x}_{n+1}}}{{{x}_{1}}}$
$\Rightarrow \sum\limits_{ciclic}{\frac{{{p}_{1}}{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{p}_{2}}{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{p}_{n}}{{\alpha }_{n}}{{a}_{n}}+{{p}_{n+1}}{{\alpha }_{n+1}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}=}$
$\sum\limits_{cyc}{\frac{{{p}_{1}}{{\alpha }_{1}}\frac{{{x}_{1}}}{{{x}_{n+1}}}+{{p}_{2}}{{\alpha }_{2}}\frac{{{x}_{2}}}{{{x}_{n+1}}}+...+{{p}_{n}}{{\alpha }_{n}}\frac{{{x}_{n}}}{{{x}_{n+1}}}+{{p}_{n+1}}{{\alpha }_{n+1}}}{{{\alpha }_{1}}\frac{{{x}_{1}}}{{{x}_{n+1}}}+{{\alpha }_{2}}\frac{{{x}_{2}}}{{{x}_{n+1}}}+...+{{\alpha }_{n}}\frac{{{x}_{n}}}{{{x}_{n+1}}}+{{\alpha }_{n+1}}}=}$
$\sum\limits_{cyc}{\frac{{{p}_{1}}{{\alpha }_{1}}{{x}_{1}}+...+{{p}_{n}}{{\alpha }_{n}}{{x}_{n}}+{{p}_{n+1}}{{\alpha }_{n+1}}{{x}_{n+1}}}{{{\alpha }_{1}}{{x}_{1}}+{{\alpha }_{2}}{{x}_{2}}+...+{{\alpha }_{n}}{{x}_{n}}+{{\alpha }_{n+1}}{{x}_{n+1}}}=}{{p}_{1}}+{{p}_{2}}+...+{{p}_{n+1}}.$
AfterMath
04.03.2013 18:57
ionbursuc you need to work on indexes to make your statements true e.g: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=523052&p=2952742#p2952742
ionbursuc
04.03.2013 19:46
AfterMath wrote: ionbursuc you need to work on indexes to make your statements true e.g: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=38&t=523052&p=2952742#p2952742 $\sum\limits_{cyc}{\frac{{{p}_{1}}{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{p}_{2}}{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{p}_{n}}{{\alpha }_{n}}{{a}_{n}}+{{p}_{n+1}}{{\alpha }_{n+1}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}=}$ $\frac{{{p}_{1}}{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{p}_{2}}{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{p}_{n}}{{\alpha }_{n}}{{a}_{n}}+{{p}_{n+1}}{{\alpha }_{n+1}}}{{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n}}+{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n}}+...+{{\alpha }_{n}}{{a}_{n}}+{{\alpha }_{n+1}}}+$ $\frac{{{p}_{1}}{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n+1}}+{{p}_{2}}{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}...{{a}_{n+1}}+...+{{p}_{n}}{{\alpha }_{n+1}}{{a}_{n+1}}+{{p}_{n+1}}{{\alpha }_{1}}}{{{\alpha }_{2}}{{a}_{2}}{{a}_{3}}...{{a}_{n+1}}+{{\alpha }_{3}}{{a}_{3}}{{a}_{4}}...{{a}_{n+1}}+...+{{\alpha }_{n+1}}{{a}_{n+1}}+{{\alpha }_{1}}}+...+$ $\frac{{{p}_{1}}{{\alpha }_{n+1}}{{a}_{n+1}}{{a}_{1}}...{{a}_{n-1}}+{{p}_{2}}{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n-1}}+...+{{p}_{n}}{{\alpha }_{n-1}}{{a}_{n-1}}+{{p}_{n+1}}{{\alpha }_{n}}}{{{\alpha }_{n+1}}{{a}_{n+1}}{{a}_{1}}...{{a}_{n-1}}+{{\alpha }_{1}}{{a}_{1}}{{a}_{2}}...{{a}_{n-1}}+...+{{\alpha }_{n-1}}{{a}_{n-1}}+{{\alpha }_{n}}}$ ${{\alpha }_{1}}\to {{\alpha }_{2}}\to {{\alpha }_{3}}...\to {{\alpha }_{n+1}}\to {{\alpha }_{1}}$ ${{a}_{1}}\to {{a}_{2}}\to {{a}_{3}}...\to {{a}_{n+1}}\to {{a}_{1}}$ (${{p}_{1}},{{p}_{2}},...,{{p}_{n+1}}$ are fixed )
v_Enhance
03.05.2014 07:14
If we put $a = \frac{x}{w}$, $b = \frac{y}{x}$, $c = \frac{z}{y}$, $d = \frac{w}{z}$ this eliminates the condition, and the expression simplifies as simply \[ \sum_{\text{cyc}} \frac{1+\frac xw + \frac yw}{1 + \frac xw + \frac yw + \frac zw} = \sum_{\text{cyc}} \frac{w+x+y}{w+x+y+z} = 3. \]
anantmudgal09
29.09.2015 12:56
The standard substitution gives a clean solution as pointed out by others. Indeed, Put $a=\frac{x}{y}$, $b=\frac{y}{z}$, $c=\frac{z}{w}$,$d=\frac{w}{x}$ and we get $E=3$.