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Let $a, \: b, \: c$ be three variables which can take any value (Real or Complex). Given that $ab + bc + ca = \frac{1}{2}$; $a + b + c = 2$; $abc = 4$. Then find the value of $$\frac{1}{ab + c - 1} + \frac{1}{bc + a - 1} + \frac{1}{ac + b - 1}$$ I always get stuck in such problems. Please give a detailed solution and also give me an approach that I can use to solve other such kind of problems.

Daniel Fischer
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Koushik Sahu
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    Put them over a common denominator. The fraction you get will have symmetric functions of $a$, $b$ and $c$ as numerator and denominator. Any symmetric function can be expressed in terms of elementary symmetric functions. – Angina Seng Oct 03 '17 at 17:43
  • You an take advantage of the fact that $1/(ab+c-1)=1/((4/c)+c-1)=c/(4+c^2-c)$. – Jean Marie Oct 03 '17 at 18:11
  • See also : https://math.stackexchange.com/questions/2209034/finding-sum-k-0n-1-dfrac-alpha-k2-alpha-k-where-alpha-k-is-prim – lab bhattacharjee Oct 03 '17 at 18:29

3 Answers3

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Hint:

Let $y=\dfrac1{ab+c-1}$

$\implies y=\dfrac 1{\dfrac12--bc-ca+c-1}=\dfrac2{2c(1-a-b)-1}=\dfrac2{2c(c-1)-1}=\dfrac2{2c^2-2c-1} $

$$\iff2y c^2-2y c-(y+2)=0\ \ \ \ (1)$$

Again $y=\dfrac1{ab+c-1}=\dfrac c{4+c^2-c}$

$$\iff c^2y-c(y+1)+4y=0\ \ \ \ (2)$$

Solve $(1),(2)$ for $c,c^2$ and use $c^2=(c)^2$ to form a cubic equation in $y$

Now apply Vieta's formula

lab bhattacharjee
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Another way is to find a cubic $p(x)$ with roots $ab+c, bc+a, ca+b$ (which is easier than the original problem) and then use simple transformations. Here using elementary symmetric polynomials and Vieta, we get $p(x) = x^3-\frac52x^2- \frac52x-\frac{65}4$.

Now get the polynomial $p_1(x)$ which has roots $ab+c-1, bc+a-1, ca+b-1$, as $p_1(x) = p(x+1) = x^3+\frac12x^2-\frac92x-\frac{81}4$

Finally the polynomial with reciprocal roots is got by reversing coefficients just so:
$p_2(x) = -\frac{81}4x^3-\frac92x^2+\frac12x+1$

So the sum of these roots is just $-\frac92 \div \frac{81}4 = -\frac29$

Macavity
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$$\sum_{cyc}\frac{1}{ab+c-1}=\frac{\sum\limits_{cyc}(ab+c-1)(ac+b-1)}{\prod\limits_{cyc}(ab+c-1)}=$$ $$=\frac{\sum\limits_{cyc}(a^2bc+a^2b+a^2c-ab-2a+1)}{a^2b^2c^2+abc-1+\sum\limits_{cyc}(a^3bc+a^2b^2-a^2bc-a^2b-a^2c+a)}=$$ $$=\frac{2\cdot4+2\cdot\frac{1}{2}-3\cdot4-\frac{1}{2}-2\cdot2+3}{4^2+4-1+4\cdot\left(2^2-2\cdot\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2-2\cdot4\cdot2-2\cdot\frac{1}{2}+3\cdot4+2}=-\frac{18}{113}.$$

Michael Rozenberg
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