Skip to main content
Lemma
Inequalities Difficulty 4

TMUA practice: Inequalities, problem 4

A TMUA-calibre inequalities problem at difficulty 4 of 5, with the full worked solution.

The question
The set of all real numbers xx satisfying x2−2x+3x2−2x+2≥2\dfrac{x^{2}-2x+3}{\sqrt{x^{2}-2x+2}}\ge2 is
A the set of all integers\text{the set of all integers}
B the set of all rational numbers\text{the set of all rational numbers}
C the set of all positive real numbers\text{the set of all positive real numbers}
D the set of all real numbers\text{the set of all real numbers}

The correct answer is highlighted. D

Worked solution

Substitute t=x2−2x+2=(x−1)2+1t=x^{2}-2x+2=(x-1)^{2}+1. Since (x−1)2≥0(x-1)^{2}\ge0, the new variable satisfies t≥1t\ge1 (so t\sqrt{t} is defined and positive). Also x2−2x+3=t+1x^{2}-2x+3=t+1, so the inequality becomes

t+1t≥2,i.e.t+1t≥2.\frac{t+1}{\sqrt{t}}\ge2, \qquad\text{i.e.}\qquad \sqrt{t}+\frac{1}{\sqrt{t}}\ge2.

This always holds. By the AM-GM inequality applied to the two positive numbers t\sqrt{t} and 1t\dfrac{1}{\sqrt{t}},

t+1t≥2t⋅1t=2,\sqrt{t}+\frac{1}{\sqrt{t}}\ge2\sqrt{\sqrt{t}\cdot\frac{1}{\sqrt{t}}}=2,

with equality when t=1t\sqrt{t}=\dfrac{1}{\sqrt{t}}, i.e. t=1t=1.

So the inequality t+1t≥2\sqrt{t}+\dfrac{1}{\sqrt{t}}\ge2 is satisfied for every t≥1t\ge1 — that is, for every real xx.

Answer: D.