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$a + b + c = 7$ and $\dfrac{1}{a+b} + \dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{7}{10}$. Find the value of $\dfrac{a}{b+c} + \dfrac{b}{c+a} + \dfrac{c}{a+b}$.

I algebraically manipulated the second equation to get:

$\dfrac{(b+c)(c+a) + (a+b)(c+a) + (a+b)(b+c)}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$

$\dfrac{bc+ab+c^2+ac+a^2+bc+ba+ab+ac+b^2+bc}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$

$\dfrac{(a+b+c)^2}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$

$\dfrac{7^2}{(a+b)(b+c)(c+a)} = \dfrac{7}{10}$

$(a+b)(b+c)(c+a) = 70$

I'm stuck after this.

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HINT:

$$\frac a{b+c}+\frac b{c+a}+\frac c{a+b}$$

$$=\frac a{b+c}+1-1+\frac b{c+a}+1-1+\frac c{a+b}+1-1$$

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

Using summation notation, $$\sum_{a,b,c} \frac a{b+c}=-3+\sum_{a,b,c} \left(\frac a{b+c}+1\right)=-3+(a+b+c)\sum_{a,b,c}\frac1{b+c}$$

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  • $\begingroup$ @AnanayAgarwal, so the sum should be $-3+7\cdot \frac7{10}=\frac{-3\cdot10+49}{10}=\frac{19}{10}$ $\endgroup$ – lab bhattacharjee Jun 26 '13 at 10:43
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    $\begingroup$ Thanks for the help. But I'm not familiar with your use of the sigma notation. I'm used to numbers written on top and on the bottom of the sigma. Can you explain what the notation means? $\endgroup$ – Gerard Jun 26 '13 at 10:47
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    $\begingroup$ @AnanayAgarwal, $\sum \frac a{b+c}=\frac a{b+c}+\frac b{c+a}+\frac c{a+b}$ $\endgroup$ – lab bhattacharjee Jun 26 '13 at 10:48
  • $\begingroup$ Thanks! So, in general, when somebody uses this notation, its just to add everything that comes after the term and looks similar? $\endgroup$ – Gerard Jun 26 '13 at 10:50
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    $\begingroup$ @AnanayAgarwal, please have a look into math.stackexchange.com/questions/280372/… and artofproblemsolving.com/Wiki/index.php/Cyclic_sum $\endgroup$ – lab bhattacharjee Jun 26 '13 at 10:53
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Here is another direct way of beginning by calculating directly from what you know:

$$(a+b+c)\left(\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}\right)=\left(\frac c{a+b}+1\right)+\left(\frac a{b+c}+1\right)+\left(\frac b{c+a}+1\right)$$

and rearranging gives you what you need.

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It is sometimes beneficial to start from the end $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=$$ $$\frac{7-b-c}{b+c}+\frac{7-a-c}{a+c}+\frac{7-a-b}{a+b}=$$ $$7\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)-3=$$ $$\frac{49}{10}-3=\frac{19}{10}$$

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