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Friends,

I encountered the following problem the other day.

  • Let us suppose that $a+b+c=1, ab+bc+ac=2$, and $abc=3$. Evaluate this expression:

$$S= \frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}.$$

I approached it thus:

The hypotheses clearly remind us of Vieta's formulas for a cubic polynomial, so I gathered that the constants could be thought as the roots of the equation

$$x^3-x^2+2x-3=0.$$

Unfortunately, the roots are not as easy to calculate by hand. I resorted to numerical approximations to 'em obtained with the aid of Mathematica and then I proceeded to evaluate the expression in question in the said CAS. I got $-4$ as the final result, but I would really like to know how it is than an answer to this question by pure ingenuity goes...

Thanks in advance for your replies!

Jamai-Con
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    You do not need to calculate the roots individually by hand. Are you familiar with the fundamental theorem of symmetric functions? $S$ can be written as a ratio of symmetric polynomials, each of which is a polynomial in $a + b + c, ab + bc + ac, abc$. – Qiaochu Yuan Jan 16 '25 at 22:51
  • Yes, I thought of that but failed in expressing $S$ in the way you mention. Could you please enter a detailed answer along these lines? – Jamai-Con Jan 16 '25 at 22:53
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    Hint: Write it as $ \sum a^2 / ( a^2 + 3)$. Do you know a cubic whose roots are $a^2, b^2 c^2$? – Calvin Lin Jan 16 '25 at 23:03
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    We have:$$S=\sum\frac a{1-b-c+bc}=\sum\frac a{(1-b)(1-c)}=\sum\frac {a(1-a)}{(1-a)(1-b)(1-c)}\ .$$The denominator is invariant w.r.t. cyclic permutations, it is $1-1+2-3=-1$. The sum $\sum a(1-a)=\sum a-\sum a^2=1-(1^2-2\cdot 2)=4$. So the result is $-4$. – dan_fulea Jan 17 '25 at 01:33
  • @Calvin Lin: Since $(a+b+c)^2= a^2+b^2+c^2+2ab+2ac+2bc$ and $(ab+ac+bc)^2=a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2$, we have that $a^2+b^2+c^2 = -3$ , $a^2b^2+a^2c^2+b^2c^2 =-2$, and $a^2b^2c^2=9$; therefore, an example of a polynomial with the roots that you mention is $p(x)=x^3+3x^2-2x-9$. – Jamai-Con Jan 17 '25 at 03:10
  • @dan_fulea: Your solution is super cool! – Jamai-Con Jan 17 '25 at 03:20
  • @Jamai-Con That works. Or you could do a root transformation $y = \sqrt{x}$, and square both sides of $ y^{3/2} + 2y^{1/2} = y + 3$ to get the same polynomial. – Calvin Lin Jan 17 '25 at 04:43

1 Answers1

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The presence of $bc$, $ac$ and $ab$ in the denominators of each fraction provides a strong motivation to attempt multiplying the numerator and denominator by the appropriate numbers $a,b$ and $c$ respectively :

$$S=\sum_{\rm cyc} \frac {x^2}{x^2+3}$$

But, we still have not gotten anything useful . Therefore, we will try ways to escape fractional expressions. Observe that, $$x^3-x^2+2x-3=0$$ yields $$x^2=x^3+2x-3$$ and $x^3+2x-3$ has a rational root $x=1$ . So, $$x^2=(x-1)(x+\color{red}{x^2+3})$$

Thus, using the factorization above, we can reach the result that works for us :

$$ \begin{align}S=\sum_{\rm cyc} \frac {x^2}{x^2+3}&=\sum_{\rm cyc}\frac {x^2}{\frac {x^2}{x-1}-x}\\ &=\sum_{\rm cyc}\frac {x^2(x-1)}{x^2-x^2+x}\\ &=\sum_{\rm cyc}(x^2-x)\\ &=\sum_{\rm cyc}\frac {3-2x}{x}\\ &=3\sum_{\rm cyc}\frac {1}{x}-6\\ &=3\cdot \frac {2}{3}-6=-4\;.\end{align} $$

$\color{red}{\text{Simpler approach}}$

It is possible to proceed with fewer calculations . Are you familiar with using Fraction Arithmetic ? It simply works as follows :

$$k=\frac {a}{b}=\frac {c}{d} \overset{\color{#c00}{b\neq d}}\iff k=\frac {a-c}{b-d}$$

Using the above arithmetic, multiplying the numerator and denominator of the fractions by $a,b,c$ respectively and then keeping in mind that $x^2+3=x^3+2x$, we have :

$$ \begin{align}\color{#c00}{\frac{x^2}{x^2+3}}&=\frac {x^2}{x^3+2x}=\color{#0a0}{\frac {x}{x^2+2}}\\ &=\frac {\color{#c00}{x^2}-\color{#0a0}{x}}{\color{#c00}{x^2+3}-\color{#0a0}{x^2-2}}\\ &=x^2-x\\ &=3\cdot\frac {1}{x}-2~.\end{align} $$

So finally,

$$S=\sum_{\rm cyc} \frac {x^2}{x^2+3}=3\sum_{\rm cyc}\frac {1}{x}-6~.$$