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

Answer:

$\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2} =1$

And on this case the samllest possible value would be 1

Step-by-step explanation:

For this case we have the following expression:

$\frac{(y-x)^2}{(y-z)(z-x)} \frac{(z-y)^2}{(z-x)(x-y)} \frac{(x-z)^2}{(x-y)(y-z)}$

And we can rewrite this expression like this:

$\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2}$

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:

$\frac{(y-x)^2}{(x-y)^2} \frac{(x-z)^2}{(z-x)^2}\frac{(z-y)^2}{(y-z)^2} =1$

And on this case the samllest possible value would be 1