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Let $a,b,c$ satisfy the equation $x^3+px^2+qx+r=0$. Is it possible to determine $\frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}$ in terms of $p,q,r$? I stumbled upon this while thinking about an inequality problem. What i could do so far is this:

We have the relations $a+b+c=-p,ab+bc+ca=q,abc=-r$ from vietas relations. So by slightly rewriting the terms we can make them a little bit symmetric like $\frac{abc}{ac+bc}$. In this manner,the numerators become constant since $abc$ is constant. But what bothers me is we will still be left with the denominators which are not completely free of $a,b$ or $c$. And evalutating them by expansion seems to be a daunting task. Is there any clever way to approach this?

madness
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1 Answers1

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Note that $\frac{ab}{a+b} = \frac{1}{\frac{1}{a}+\frac{1}{b}}$.

Suppose we have the monic cubic $f(x) = x^3+a_2x^2+a_1x+a_0$ with roots $\alpha, \beta, \gamma$. We can make use of the following two operations:

  1. $\frac{1}{a_0}x^3f(\frac{1}{x})$ is the unique monic cubic with roots $\frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}$.
  2. $-f(\alpha+\beta+\gamma-x) = -f(-a_2-x)$ is the unique monic cubic with roots $\alpha+\beta,\,\alpha+\gamma,\,\beta+\gamma $.

Applying (1) to the original cubic: $x^3+\frac{q}{r}x^2+\frac{p}{r}x+\frac{1}{r}$ has roots $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$.

Applying (2) to the result: $$\left(x+\tfrac{q}{r}\right)^3 - \tfrac{q}{r}\left(x+\tfrac{q}{r}\right)^2 +\tfrac{p}{r}\left(x+\tfrac{q}{r}\right) - \tfrac{1}{r} = x^3+\tfrac{2q}{r}x^2+\tfrac{q^2+pr}{r^2} x + \tfrac{pq-r}{r^2}$$ has roots $\frac{1}{a}+\frac{1}{b},\, \frac{1}{a}+\frac{1}{c},\, \frac{1}{b}+\frac{1}{c}$.

Finally, applying (1) again gives $$x^3+\tfrac{q^2+pr}{pq-r}x^2 + \tfrac{2qr}{pq-r}x+\tfrac{r^2}{pq-r}$$ which has roots $\frac{1}{\frac{1}{a}+\frac{1}{b}},\,\frac{1}{\frac{1}{a}+\frac{1}{c}},\,\frac{1}{\frac{1}{b}+\frac{1}{c}}$. Your cyclic sum is the sum of the roots of this polynomial, which by Vieta is $$-\frac{q^2+pr}{pq-r}.$$

Benjamin Wright
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