Where's the other 4 equations?

Algebra Level 5

Let $a, b, c, x, y, z$ be positive real numbers such that

$\begin{aligned} abx + bcy + acz &= xyz - 2abc \\ axz + bxy + cyz &= xyz. \\ \end{aligned}$

How many solutions are there to this system?

If there are an infinite number of solutions, input $-1$ as your answer. If there are no solutions to the system, input $0$ .

1 solution

Steven Yuan
Apr 15, 2015

The first equation can be rewritten as

$\frac{ab}{yz} + \frac{bc}{zx} + \frac{ca}{xy} + 2 \frac{abc}{yzx} = 1.$

This shows the existence of positive real numbers $u, v, w$ such that

$\begin{aligned} \frac{a}{y} &= \frac{u}{v + w} \\ \frac{b}{z} &= \frac{v}{u + w} \\ \frac{c}{x} &= \frac{w}{u + v}. \end{aligned}$

The second equation says that $\frac{a}{y} + \frac{b}{z} + \frac{c}{x} = 1.$ However, from Nesbitt's Inequality, we find that

$1 = \frac{a}{y} + \frac{b}{z} + \frac{c}{x} = \frac{u}{v + w} + \frac{v}{u + w} + \frac{w}{u + v} \geq \frac{3}{2},$

a contradiction. Thus, there are $\boxed{0}$ solutions to the system of equations.