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 4 of 5: Parametrize the equality system a+b+c=1, ab+bc+ca=0 by a rational t
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In plain words

A one-parameter rational curve of solutions is exactly what is needed for infinitely many rational triples, so we look for one by introducing a rational slope t = b/a.

a=1+tt2+t+1, b=t(1+t)t2+t+1, c=−tt2+t+1a=\frac{1+t}{t^2+t+1},\ b=\frac{t(1+t)}{t^2+t+1},\ c=\frac{-t}{t^2+t+1}
Detailed analysis

Seek solutions of a+b+c=1a+b+c=1, ab+bc+ca=0ab+bc+ca=0 with b=tab=ta for rational tt. Substituting c=1−a−b=1−a(1+t)c=1-a-b=1-a(1+t) into ab+bc+ca=0ab+bc+ca=0 and dividing by a≠0a\ne0 gives a[(1+t)2−t]=1+ta\left[(1+t)^2-t\right]=1+t, i.e.\ a=1+tt2+t+1a=\frac{1+t}{t^2+t+1} (note t2+t+1>0t^2+t+1>0 always). Then b=ta=t(1+t)t2+t+1b=ta=\frac{t(1+t)}{t^2+t+1}, and c=1−a−b=(t2+t+1)−(1+t)−t(1+t)t2+t+1=−tt2+t+1c=1-a-b=\frac{(t^2+t+1)-(1+t)-t(1+t)}{t^2+t+1}=\frac{-t}{t^2+t+1}.