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 1 of 5: Substitute a = x/(x-1) to turn the target sum into a squared sum
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a=xx−1, b=yy−1, c=zz−1  ⟹  x=aa−1, y=bb−1, z=cc−1a=\frac{x}{x-1},\ b=\frac{y}{y-1},\ c=\frac{z}{z-1} \implies x=\frac{a}{a-1},\ y=\frac{b}{b-1},\ z=\frac{c}{c-1}
Detailed analysis

Define a=xx−1a=\frac{x}{x-1}, b=yy−1b=\frac{y}{y-1}, c=zz−1c=\frac{z}{z-1}; the target sum is exactly a2+b2+c2a^2+b^2+c^2. Each substitution is invertible: from a=xx−1a=\frac{x}{x-1} we get a(x−1)=xa(x-1)=x, i.e.\ x(a−1)=ax(a-1)=a, so x=aa−1x=\frac{a}{a-1} (and likewise for y,zy,z); in particular a=1a=1 would force x=x−1x=x-1, impossible, so a,b,c≠1a,b,c\ne1 automatically.