Solve Algebra Challenge: Find $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$

[anemone]

If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$.

 

[anemone]

Hint:

Spoiler

Let $a=x+y,\,b=y+z,\,c=z+x$ and see where that leads you...

 

[Albert1]

the answer is:

Spoiler

$\dfrac {4031}{4030}$
correct ?

 

[anemone]

Spoiler

That is $\text{Correct}$.

 

[Albert1]

If $\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$, evaluate $\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}----(1)$.

Spoiler

let $x\geq 0, y\geq 0, z\geq 0$
and let :$a=\dfrac {x-y}{x+y}, b=\dfrac {y-z}{y+z}, c=\dfrac {z-x}{z+x}$
then $abc=\dfrac {2014}{2015}=-\dfrac {2014}{2015}\times 1\times(-1)$
$a,b,c$ must be 1 positive and 2 negatives (since 3 positives are impossible)
by observation if $z=0 $ then $c=-1<0, b=1>0($ independent of $x$ and $y$)
and $a=-\dfrac {2014}{ 2015}<0$
so we only have to find the values of $x$ and $y$
that is $x+y=2015---(2),x-y=-2014---(3)$
from (2)(3)$\therefore x=\dfrac {1}{2},y=\dfrac {4029}{2}$
and (1) becomes :
$\dfrac{\dfrac{1}{2}}{2015}+1+0=\dfrac {1}{4030}+1=\dfrac {4031}{4030}$

 

[anemone]

Spoiler

...
by observation if $z=0 $...

Hi again Albert,

Spoiler

I see that if you let $z=0$, then our target expression becomes undefined.

 

[Albert1]

Hi again Albert,

Spoiler

I see that if you let $z=0$, then our target expression becomes undefined.

Spoiler

why ? Our target expression becomes undefined ?

take note my definition of $a,b,c$

 

[anemone]

why ? our target expression becomes undefined ?

take note my definition of $a,b,c$

Hi Albert,

Spoiler

Ops, I don't know where my head was today, I made a wrong assumption and replied hastily to your solution without thinking for more. I am sorry.

 

[maxkor]

Hint:

Spoiler

Let $a=x+y,\,b=y+z,\,c=z+x$ and see where that leads you...

Show your solution

 

[MarkFL]

Show your solution

It's only been 4 days since anemone posted a hint to the problem. She is trying to judiciously encourage participation in this problem. So, let's give her some more time before we command her to post the solution, okay?

 

[anemone]

Show your solution

I'm thinking perhaps the previous statement is a not-so-subtle hint, I will post for further hints before revealing the solution, okay?

Hint 1:

Spoiler

Let $a=x+y,\,b=y+z,\,c=z+x$ and see where that leads you...

Hint 2:

Spoiler

Note that $a-c=x+y-z-x=y-z$, $b-a=y+z-x-y=z-x$ and $c-b=z+x-y-z=x-y$

Also $a-c+b=2y$, $b-a+c=2z$ and $c-b+a=2x$

Hint 3:

Spoiler

$(c-b)(a-c)(b-a)=a^2b-a^2c-ab^2+ac^2+b^2c-bc^2$

 

[anemone]

My solution:

Spoiler

Let $a=x+y,\,b=y+z,\,c=z+x$, note that

$a-c=x+y-z-x=y-z$,

$b-a=y+z-x-y=z-x$ and

$c-b=z+x-y-z=x-y$.

Also,

$a-c+b=2y$,

$b-a+c=2z$, and

$c-b+a=2x$.

Therefore, the given equation becomes:$\dfrac{(x-y)(y-z)(z-x)}{(x+y)(y+z)(z+x)}=\dfrac{2014}{2015}$

$\dfrac{(c-b)(a-c)(b-a)}{(a)(b)(c)}=\dfrac{2014}{2015}$And our target expression becomes:

$\dfrac{x}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{z+x}$

$=\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}$

$=\dfrac{c-b+a}{2a}+\dfrac{a-c+b}{2b}+\dfrac{b-a+c}{2c}$

$=\dfrac{a}{2a}+\dfrac{b}{2b}+\dfrac{c}{2c}+\dfrac{c-b}{2a}+\dfrac{a-c}{2b}+\dfrac{b-a}{2c}$

$=\dfrac{3}{2}+\dfrac{1}{2}\left(\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}\right)$

$=\dfrac{3}{2}+\dfrac{1}{2}\left(-\dfrac{(c-b)(a-c)(b-a)}{abc}\right)$

$=\dfrac{3}{2}-\dfrac{1}{2}\left(\dfrac{2014}{2015}\right)$

$=\dfrac{4031}{4030}$

since $\dfrac{c-b}{a}+\dfrac{a-c}{b}+\dfrac{b-a}{c}=-\dfrac{(c-b)(a-c)(b-a)}{abc}$.