Maths Olympiad Prep

Track / Stage 7 / 11 of 300 #1411 of 1964

Problem 1411

National Olympiad second round; IMO P1/P4
Algebra Difficulty 7.0 Prove it

Let x,y,z x, y, z be distinct real numbers. Prove that
x2(x−y)2+y2(y−z)2+z2(z−x)2≥1 \frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Next problem →

Official solution

Solution. We have
∑cyc(1−xz)2(1−zy)2−(1−yx)2(1−zy)2(1−xz)2==∑cyc(1−xz−zy+xy)2−(xy+yz+zx−yx−zy−xz)2==3+2∑cycx2z2+∑cycx2y2−4∑cycxz+4∑cycxy−2∑cycx2yz−2∑cyc2yzx2−∑cycx2y2−∑cycy2x2+2(∑cycxy)(∑cycyx)−2∑cycyx−2∑cycxy==∑cycx2z2−6∑cycxz+2∑cycxy+9=(xz+zy+yx−3)2≥0.\begin{array}{c} \sum_{c y c}\left(1-\frac{x}{z}\right)^{2}\left(1-\frac{z}{y}\right)^{2}-\left(1-\frac{y}{x}\right)^{2}\left(1-\frac{z}{y}\right)^{2}\left(1-\frac{x}{z}\right)^{2}= \\ =\sum_{c y c}\left(1-\frac{x}{z}-\frac{z}{y}+\frac{x}{y}\right)^{2}-\left(\frac{x}{y}+\frac{y}{z}+\frac{z}{x}-\frac{y}{x}-\frac{z}{y}-\frac{x}{z}\right)^{2}= \\ =3+2 \sum_{c y c} \frac{x^{2}}{z^{2}}+\sum_{c y c} \frac{x^{2}}{y^{2}}-4 \sum_{c y c} \frac{x}{z}+4 \sum_{c y c} \frac{x}{y}-2 \sum_{c y c} \frac{x^{2}}{y z}-2 \sum_{c y c} 2 \frac{y z}{x^{2}} \\ -\sum_{c y c} \frac{x^{2}}{y^{2}}-\sum_{c y c} \frac{y^{2}}{x^{2}}+2\left(\sum_{c y c} \frac{x}{y}\right)\left(\sum_{c y c} \frac{y}{x}\right)-2 \sum_{c y c} \frac{y}{x}-2 \sum_{c y c} \frac{x}{y}= \\ =\sum_{c y c} \frac{x^{2}}{z^{2}}-6 \sum_{c y c} \frac{x}{z}+2 \sum_{c y c} \frac{x}{y}+9 \\ =\left(\frac{x}{z}+\frac{z}{y}+\frac{y}{x}-3\right)^{2} \geq 0 . \end{array}

We conclude that
∑cyc(1−xz)2(1−zy)2≥(1−yx)2(1−zy)2(1−xz)2⇒⇒∑cycx2(z−x)2(z−y)2≥(x−y)2(y−z)2(z−x)2⇒⇒x2(x−y)2+y2(y−z)2+z2(z−x)2≥1\begin{array}{c} \sum_{c y c}\left(1-\frac{x}{z}\right)^{2}\left(1-\frac{z}{y}\right)^{2} \geq\left(1-\frac{y}{x}\right)^{2}\left(1-\frac{z}{y}\right)^{2}\left(1-\frac{x}{z}\right)^{2} \Rightarrow \\ \Rightarrow \sum_{c y c} x^{2}(z-x)^{2}(z-y)^{2} \geq(x-y)^{2}(y-z)^{2}(z-x)^{2} \Rightarrow \\ \Rightarrow \frac{x^{2}}{(x-y)^{2}}+\frac{y^{2}}{(y-z)^{2}}+\frac{z^{2}}{(z-x)^{2}} \geq 1 \end{array}
as desired. Equality holds for all triples (x,y,z)(x, y, z) such that xz+zy+yx=3\frac{x}{z}+\frac{z}{y}+\frac{y}{x}=3.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.