Find the value of abc

Question

Suppose $a$, $b$, $c$, are real numbers such that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}=4$$ Determine the value of abc.

I've encountered this problem as I was going over some contest papers for practice. Quite frankly, I'm lost on how to get the numerical value of $abc$ from the equations. Using systems, I just go around in circles.

Looking for an example solution to put in my notes. Thank you very much to whoever can.

Answer 1

$$a+b+c=(a+b+c)\sum_{cyc}\frac{1}{a+b}=\sum_{cyc}\left(1+\frac{c}{a+b}\right)=3+4=7.$$ The rest is smooth: $$\sum_{cyc}(a^2+3ab)=\prod_{cyc}(a+b)$$ or $$(a+b+c)^2+ab+ac+bc=(a+b+c)(ab+ac+bc)-abc$$ or $$(a+b+c)^2+4abc=4(a+b+c)abc-abc.$$

I got $$abc=\frac{49}{23}.$$

Answer 2

Write the equations as $$ \eqalign{{a}^{-1}+{b}^{-1}+{c}^{-1}-4&=0\cr \left( a+b \right) ^{-1}+ \left( b+c \right) ^{-1}+ \left( c+a \right) ^{-1}-1&=0\cr{\frac {c}{a+b}}+{\frac {a }{b+c}}+{\frac {b}{c+a}}&-4=0}$$

By solving for $c$ in the first and substituting into the other two equations, then taking the resultant of the numerators with respect to $b$, I get $$ {a}^{24} \left( 4\,a+1 \right) ^{2} \left( 23\,{a}^{3}-161\,{a}^{2}+ 196\,a-49 \right) ^{2} \left( 4\,a-1 \right) ^{8}=0 $$ $a=0$ is impossible as $1/a$ wouldn't be defined. $a=1/4$ wouldn't work as you get $b=-c$, in the first equation, then $(b+c)^{-1}$ wouldn't be defined. $a=-1/4$ turns out not to work either. So we must have $P(a) = 23 a^3 - 161 a^2 + 196 a - 49=0$. $P$ has three roots. By symmetry (if $(a,b,c)$ is a solution then so are $(b,c,a)$ and $(c,a,b)$), $b$ and $c$ are also roots of this polynomial; they must be different roots, since with $a=b$ or $a=c$ or $b=c$ the equations don't have a solution. Thus $abc$ is the product of the three roots, which is $49/23$.