Find the smallest possible value of(y−x)2(y−z)(z−x)+(z−y)2(z−x)(x−y)+(x−z)2(x−y)(y−z),where x,y, and z are distinct real numbers.
+506 You could apply AM-GM to get
(y−x)2(y−z)(z−x)+(z−y)2(z−x)(x−y)+(x−z)2(x−y)(y−z)≥33√(y−x)2(y−z)(z−x)⋅(z−y)2(z−x)(x−y)⋅(x−z)2(x−y)(y−z)=3
but idk if that's correct because it says real numbers not positive reals
+118703
(y−x)2(y−z)(z−x)+(z−y)2(z−x)(x−y)+(x−z)2(x−y)(y−z)=(y−x)2(x−y)(y−z)(z−x)(x−y)+(z−y)2(y−z)(z−x)(x−y)(y−z)+(x−z)2(z−x)(x−y)(y−z)(z−x)=[(y−x)2(x−y)]+[(z−y)2(y−z)]+[(x−z)2(z−x)](y−z)(z−x)(x−y)=[−(y−x)2(y−x)]−[(z−y)2(z−y)]−[(x−z)2(x−z)](yz−yx−z2+xz)(x−y)=−{[(y−x)3]+[(z−y)3]+[(x−z)3]}(yz−yx−z2+xz)(x−y)=−{[(y−x)3]+[(z−y)3]+[(x−z)3]}(yzx−yzy−yxx+yxy−z2x+z2y+xzx−xzy)=−{[y3−3y2x+3yx2−x3]+[z3−3z2y+3zy2−y3]+[x3−3x2z+3xz2−z3]}(−y2z−x2y+y2x−z2x+z2y+x2z)=−3{[−y2x+yx2]+[−z2y+zy2]+[−x2z+xz2]}(−y2z−x2y+y2x−z2x+z2y+x2z)=−3(zy2+yx2−y2x+xz2−z2y−x2z(−y2z−x2y+y2x−z2x+z2y+x2z)=3so 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}
