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I just stumbled upon a contest question from last year's city olympiad math contest:

Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.

Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.

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

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Multiplying the given expressions together:

\begin{align} \frac{47}{10} &= (a+b+c)\bigg(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\bigg) \\ \\ &= \frac{a+b+c}{a+b}+\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a} \\ \\ &=3+\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} \\ \\ \end{align}

$$\Longrightarrow \frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a} = \frac{17}{10}$$

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$$a+b+c=6\tag{1}$$

$$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}=\frac{47}{60}\tag{2}$$

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

$S = \left(\frac{a+b+c}{b+c}-1\right)+\left(\frac{a+b+c}{c+a}-1\right)+\left(\frac{a+b+c}{a+b}-1\right) =6\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3=6\cdot\frac{47}{60}-3$

$S =\frac{47}{10}-3 =\frac{17}{10}$

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  • $\begingroup$ Beat me to it. Typing slow on my mobile device. $\endgroup$ – Oscar Lanzi Jun 7 '17 at 1:22
  • $\begingroup$ @OscarLanzi It's strange this answer gets so many up votes given that the content is very simple. $\endgroup$ – mrnovice Jun 7 '17 at 23:59
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Hint:$$\frac{a+b+c}{b+c}=1+\frac{a}{b+c}$$

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Even if the Joshua's answer is probably the most elegant solution, let me offer you an alternative way of solving it, that is through simplification due to arbitrary values.

You have three unknowns and only two equations. This allows you to get rid of one variable, for example, c = 0. (Note that only one can be zero).

Hence, the problem can be rewritten as:

1) $a + b = 6$

2) $\dfrac{1}{a + b} + \dfrac{1}{b} + \dfrac{1}{a} = \dfrac{1}{a + b} + \dfrac{a + b}{ab} = \dfrac{47}{60}$

$S = \dfrac{a}{b} + \dfrac{b}{a} = \dfrac{a^2 + b^2}{ab} = \dfrac{(a + b)^2 - 2ab}{ab}$.

At this point, you can clearly either find both the values of $a$ and $b$ (annoying) or trick a little more.

Substituting $x = a + b$ and $y = ab$, we have the perfectly equivalent problem

1) $x = 6$

2) $\dfrac{1}{x} + \dfrac{x}{y} = \dfrac{47}{60}$

$S = \dfrac{x^2 - 2y}{y}$.

Substituting the first equation to the second, we get $\dfrac{6}{y} = \dfrac{47}{60} - \dfrac{1}{6} = \dfrac{37}{60}$ and $y = \dfrac{10\cdot6^2}{37}$.

Finally, $S = \dfrac{x^2 - 2y}{y} = \left(6^2-2\cdot\dfrac{10\cdot6^2}{37}\right)\dfrac{37}{10\cdot6^2} = \left(1-\dfrac{20}{37}\right)\dfrac{37}{10} = \dfrac{37-20}{37} \cdot \dfrac{37}{10} = \dfrac{17}{10}$.

Answer $S = \dfrac{17}{10}$.

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  • $\begingroup$ Great first answer! Well done! $\endgroup$ – user370967 Jun 7 '17 at 18:35
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The only possible numbers a, b, c that add up to 6 are 1, 2, 3.

Substituting these in the first equation gives you 47/60 so they are the correct values and there is no repeated number for example a and b both being 1 and c being 4.

So simply substituting in the second equation gives you 17/10 immediately.

The three fractions in each equation are all the available permutations.

There is no need for any other manipulations.

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    $\begingroup$ $a,b,c\in\mathbb{R}$ $\endgroup$ – Foobaz John Jun 7 '17 at 16:07
  • $\begingroup$ You have made an unsupported assumption that a,b,c are naturals. That it (and that further they are distinct) is correct makes the problem a little silly, but doesn't mean your method is good. If you have a non-guess and check way to quickly find a,b,c it would be. $\endgroup$ – user321537 Jun 7 '17 at 17:19
  • $\begingroup$ If you look at the first equation then of course any real numbers a, b, c can sum to 6. But you must constrain yourself by the requirement introduced by the second equation. Solutions based on rearranging equation (2) to find equation (1) as a factor - and then substitute 47/60 - are correct - but not for irrationals or complex numbers. However if you substitute a negative for a, b, c the the problem becomes: a + b - c = 6 5, 3, -2, satisfy equation 1, but not equation 2. etc etc. $\endgroup$ – Not Disclosed Jun 9 '17 at 8:27

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