Can You Solve This Unique Fraction Challenge?

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SUMMARY

The forum discussion centers on evaluating the expression $\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}$ under the condition that $\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\dfrac{1}{11}$. Members maxkor and DaalChawal successfully solved the problem, confirming that the unique fraction challenge can be approached through algebraic manipulation and substitution techniques. The solution highlights the importance of understanding relationships between variables in fractional expressions.

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Here is this week's POTW:

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Evaluate $\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}$, given that $\dfrac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}=\dfrac{1}{11}$.

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Congratulations to the following members for their correct solution! (Cool)
1. maxkor
2. DaalChawal

Solution from DaalChawal:
We have
$a \over a+b$ + $b \over b+c$ +$c \over c+a$

= $1 \over 2$ $[3 + \sum\limits_{cyc} \frac{(a-b)}{(a+b)} ]$

= $\frac{1}{2}$ $[3 - \prod\limits_{cyc} \frac{(a-b)}{(a+b)}]$

= $\frac{1}{2} [3 - \frac{1}{11}]$

= $\frac{16}{11}$
 

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