Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.4 National Olympiad Prove it Belarus

Find all possible values of the expression
$\frac{1}{a}\left(\frac{1}{b}+\frac{1}{c}+\frac{1}{b+c}\right)+\frac{1}{b}\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{c+a}\right)+\frac{1}{c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{a+b}\right)-\frac{1}{a+b+c}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
if positive numbers $a, b, c$ satisfy the condition $ab + bc + ca = abc$ . (S. Chernov)

Solution

Answer: $1$ .
By condition, $1/a + 1/b + 1/c = 1$ . Let $x = 1/a$ , $y = 1/b$ , $z = 1/c$ . Then
$x + y + z = 1. \quad (1)$
We rewrite the initial expression as
$\begin{align*} &+ \left( \frac{xyz}{x+y} - \frac{xyz}{xy+yz+zx} \cdot \frac{xy}{x+y} \right) - \frac{xyz}{xy+yz+zx} = \\ &= 1 + \frac{xyz(xy + zx)}{(xy + yz + zx)(y + z)} + \frac{xyz(xz + yz)}{(xy + yz + zx)(x + y)} + \\ &\qquad + \frac{xyz(zy + xy)}{(xy + yz + zx)(z + x)} - \frac{xyz}{xy + yz + zx} = \\ &= 1 + \frac{x^2yz}{xy + yz + zx} + \frac{xyz^2}{xy + yz + zx} + \frac{xy^2z}{xy + yz + zx} - \frac{xyz}{xy + yz + zx} = \\ &= \therefore 1 + \frac{xyz(x + y + z)}{xy + yz + zx} - \frac{xyz}{xy + yz + zx} = 1. \end{align*}$

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