Question

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.

Answer 1

Combining all three fractions under a single denominator, the given expression is equal to Consider the numerator as a polynomial in , so that (where we treat and as fixed values). It follows that , so is a root of and divides into . By symmetry, it follows that and divide into . Since is a cubic in its variables, it follows that , where is a constant. By either expanding the definition of , or by trying test values (if we take , we obtain ), it follows that . Thus, Solution 2:Combining all three fractions under a single denominator, the given expression is equal to Using the sum of three cubes identity which can be proved by direct expansion, it follows that if . Letting ,then , so Solution 3:We can re-write the expression as Let us consider the first term in the sum. Expanding the numerator, it follows that By symmetry, it follows that the original expression is equal to The terms share a common denominator and furthermore sum to By symmetry, the desired answer is 3.

Answer 2

`((y-x)^2)/((x-y)^2) ((x-z)^2)/((z-x)^2)((z-y)^2)/((y-z)^2) =1`

And on this case the samllest possible value would be 1

Explanation:

For this case we have the following expression:

`((y-x)^2)/((y-z)(z-x)) ((z-y)^2)/((z-x)(x-y)) ((x-z)^2)/((x-y)(y-z))`

And we can rewrite this expression like this:

`((y-x)^2)/((x-y)^2) ((x-z)^2)/((z-x)^2)((z-y)^2)/((y-z)^2)`

For this case is important to remember the following property from algebra:

`(a-b)^2 = a^2 -2ab + b^2`

`(b-a)^2 = b^2 - 2ab + a^2`

On this case we can see that
`(a-b)^2 = (b-a)^2`

So then
`(y-x)^2 = (x-y)^2 , (x-z)^2= (z-x)^2, (z-y)^2 =(y-z)^2`

So then we can simplify all the expression and we got this:

`((y-x)^2)/((x-y)^2) ((x-z)^2)/((z-x)^2)((z-y)^2)/((y-z)^2) =1`

And on this case the samllest possible value would be 1