MathLabs

Problem 2

Let xx, yy, zz be real numbers, all different from 11, such that xyz=1xyz=1. Prove that x2(x−1)2+y2(y−1)2+z2(z−1)2≥1,\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1, and prove that equality holds for infinitely many triples of rational numbers xx, yy, zz.
Step 5 of 5: Convert back to x, y, z: an explicit infinite rational equality family
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(x,y,z)=(−1+tt2, −t(1+t), t(t+1)2),t∈Q∖{0,−1}(x,y,z)=\left(-\frac{1+t}{t^2},\ -t(1+t),\ \frac{t}{(t+1)^2}\right),\quad t\in\mathbb{Q}\setminus\{0,-1\}
Detailed analysis

Using x=aa−1x=\frac{a}{a-1} with a−1=−t2t2+t+1a-1=\frac{-t^2}{t^2+t+1} gives x=−1+tt2x=-\frac{1+t}{t^2}; using y=bb−1y=\frac{b}{b-1} with b−1=−1t2+t+1b-1=\frac{-1}{t^2+t+1} gives y=−t(1+t)y=-t(1+t); using z=cc−1z=\frac{c}{c-1} with c−1=−(t+1)2t2+t+1c-1=\frac{-(t+1)^2}{t^2+t+1} gives z=t(t+1)2z=\frac{t}{(t+1)^2}. A direct check gives xyz=1xyz=1 for every rational t≠0,−1t\ne0,-1, and since t2+t+1=0t^2+t+1=0 has no real root, none of x,y,zx,y,z ever equals 11. As tt ranges over the infinitely many rationals other than 0,−10,-1, the value y=−t(1+t)y=-t(1+t) takes infinitely many distinct values, so this produces infinitely many rational triples (x,y,z)(x,y,z) achieving equality, proving part (ii). ■\blacksquare