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The result below has been disproven.

Prove that the equation $$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} = 6$$ has no solutions in natural numbers.

The equation is equivalent to $$(a+b+c)\left(\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\right) = 9.$$ Then since $\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b} \leq \dfrac{3}{2}$, it follows that $a+b+c \geq 6$. Suppose without loss of generality that $a \leq b \leq c$. Then $$\dfrac{c}{2a}+\dfrac{c}{2a}+\dfrac{c}{2a} = \dfrac{3c}{2a} \geq 6.$$ Thus $c \geq 4a$. How can I continue?

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There is a solution with $n = 6$. See this paper (An unusual cubic representation problem by Andrew Bremner and Allan MacLeod) as well as Micheal Stoll's answer on MathOverflow.

The equation

$$ \frac{a}{a + b} + \frac{b}{a + c} + \frac{c}{a + b} = 6 $$

corresponds to the equation

$$ a^3 + b^3 + c^3 + abc - 5(a + b)(b + c)(a + c) = 0$$

which gives a projective cubic curve. If we fix a point on the curve to be the origin, we get the following elliptic curve over $\mathbf{Q}$:

$$ E: y^2 = x(x^2 + 213 x + 288). $$

You can use this information to compute $E(\mathbf{Q})$ and obtain solutions to the equation. In their paper, Bremmer and Macleod state that for $n = 6$ the minimum solution has 134 digits for the largest of $a, b, c$.

One solution is found here. (Thanks to Oleg567).

a=1218343242702905855792264237868803223073090298310121297526752830558323845503910071851999217959704024280699759290559009162035102974023;

b=2250324022012683866886426461942494811141200084921223218461967377588564477616220767789632257358521952443049813799712386367623925971447;

c=20260869859883222379931520298326390700152988332214525711323500132179943287700005601210288797153868533207131302477269470450828233936557.

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