Fractional Inequality
Find the smallest possible value of $$\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)},$$ where $ x,y, $ and $ z $ are distinct real numbers.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Combining all three fractions under a single denominator, the given expression is equal to $$\frac{(x-y)^3 + (y-z)^3 + (z-x)^3}{(x-y)(y-z)(z-x)}.$$ Consider the numerator as a polynomial in $ x $, so that $ P(x) = (x-y)^3 + (y-z)^3 + (z-x)^3 $ (where we treat $ y $ and $ z $ as fixed values). It follows that $ P(y) = (y-y)^3 + (y-z)^3 + (z-y)^3 = 0 $, so $ y $ is a root of $ P(x) = 0 $ and $ x-y $ divides into $ P(x) $. By symmetry, it follows that $ y-z $ and $ z-x $ divide into $ P(x) $. Since $ P $ is a cubic in its variables, it follows that $ P = k(x-y)(y-z)(z-x) $, where $ k $ is a constant. By either expanding the definition of $ P $, or by trying test values (if we take $ x = 0, y = -1, z = 1 $, we obtain $ P = -6 = k \cdot (-2) $), it follows that $ k = 3 $. Thus, $$\frac{(x-y)^3 + (y-z)^3 + (z-x)^3}{(x-y)(y-z)(z-x)} = \boxed{3}.$$
Freeflows