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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.

 Feb 18, 2022
 #1
avatar+499 
+1

You could apply AM-GM to get

\(\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\ge3 \sqrt[3]{\frac{(y-x)^2}{(y-z)(z-x)} \cdot\frac{(z-y)^2}{(z-x)(x-y)} \cdot\frac{(x-z)^2}{(x-y)(y-z)}}\\=\boxed{3}\)

but idk if that's correct because it says real numbers not positive reals

 Feb 19, 2022
 #2
avatar+117834 
+1

 

\(\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\\ =\frac{(y-x)^2(x-y)}{(y-z)(z-x)(x-y)} + \frac{(z-y)^2(y-z)}{(z-x)(x-y)(y-z)} + \frac{(x-z)^2(z-x)}{(x-y)(y-z)(z-x)}\\\\ =\frac{[(y-x)^2(x-y)]\;\;+\;\;[(z-y)^2(y-z)]\;\; +\;\; [(x-z)^2(z-x)]}{(y-z)(z-x)(x-y)} \\ =\frac{[-(y-x)^2(y-x)]\;\;-\;\;[(z-y)^2(z-y)]\;\; -\;\; [(x-z)^2(x-z)]}{(yz-yx-z^2+xz)(x-y)} \\ =\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\; +\;\; [(x-z)^3]\}}{(yz-yx-z^2+xz)(x-y)} \\ =\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\; +\;\; [(x-z)^3]\}}{(yzx-yzy-yxx+yxy-z^2x+z^2y+xzx-xzy)} \\ =\frac{-\{[y^3-3y^2x+3yx^2-x^3]\;\;+\;\;[z^3-3z^2y+3zy^2-y^3]\;\; +\;\; [x^3-3x^2z+3xz^2-z^3]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =\frac{-3\{[-y^2x+yx^2]\;\;+\;\;[-z^2y+zy^2]\;\; +\;\; [-x^2z+xz^2]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =\frac{-3(zy^2+yx^2 -y^2x+xz^2-z^2y-x^2z}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\ =3 \qquad \text{so long as the denominator isn't 0}\)

 

I'm sure Textot's formula is what you were supposed to use.

 

 

LaTex

\frac{(y-x)^2}{(y-z)(z-x)} + \frac{(z-y)^2}{(z-x)(x-y)} + \frac{(x-z)^2}{(x-y)(y-z)}\\
=\frac{(y-x)^2(x-y)}{(y-z)(z-x)(x-y)} + \frac{(z-y)^2(y-z)}{(z-x)(x-y)(y-z)} + \frac{(x-z)^2(z-x)}{(x-y)(y-z)(z-x)}\\\\
=\frac{[(y-x)^2(x-y)]\;\;+\;\;[(z-y)^2(y-z)]\;\;  +\;\;  [(x-z)^2(z-x)]}{(y-z)(z-x)(x-y)} \\
=\frac{[-(y-x)^2(y-x)]\;\;-\;\;[(z-y)^2(z-y)]\;\;  -\;\;  [(x-z)^2(x-z)]}{(yz-yx-z^2+xz)(x-y)} \\
=\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\;  +\;\;  [(x-z)^3]\}}{(yz-yx-z^2+xz)(x-y)} \\
=\frac{-\{[(y-x)^3]\;\;+\;\;[(z-y)^3]\;\;  +\;\;  [(x-z)^3]\}}{(yzx-yzy-yxx+yxy-z^2x+z^2y+xzx-xzy)} \\
=\frac{-\{[y^3-3y^2x+3yx^2-x^3]\;\;+\;\;[z^3-3z^2y+3zy^2-y^3]\;\;  +\;\;  [x^3-3x^2z+3xz^2-z^3]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=\frac{-3\{[-y^2x+yx^2]\;\;+\;\;[-z^2y+zy^2]\;\;  +\;\;  [-x^2z+xz^2]\}}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=\frac{-3(zy^2+yx^2 -y^2x+xz^2-z^2y-x^2z}{(-y^2z-x^2y+y^2x-z^2x+z^2y+x^2z)} \\
=3  \qquad \text{so long as the denominator isn't 0}

 Feb 19, 2022
 #3
avatar+499 
0

Thanks melody, I didn't realize the whole thing reduced to 3 haha

Also, if you were to go to WolframAlpha and try to minimize the function, it gives the correct minimum value but at a strangely specific point:

https://www.wolframalpha.com/input?i=minimize+%5Cfrac%7B%28y-x%29%5E2%7D%7B%28y-z%29%28z-x%29%7D+%2B+%5Cfrac%7B%28z-y%29%5E2%7D%7B%28z-x%29%28x-y%29%7D+%2B+%5Cfrac%7B%28x-z%29%5E2%7D%7B%28x-y%29%28y-z%29%7D%2C

Don't know why that is.

textot  Feb 19, 2022
 #4
avatar+117834 
0

I don't know either ......

Melody  Feb 19, 2022

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