Maths WorkshopMaths problems with full solutions
Exercise 1991Algebra › InequalitiesOlympiadAdded 5 Nov 2026

With abc = 1: Σ 1/(a² + a + 1) ≥ 1

#constraint abc=1#clearing denominators#expansion

Statement

Let a,b,c>0a,b,c>0 with abc=1abc=1. Prove that
1a2+a+1+1b2+b+1+1c2+c+1≥1.\frac1{a^2+a+1}+\frac1{b^2+b+1}+\frac1{c^2+c+1}\ge1.

Hint

Clear the denominators and use abc=1abc=1 to simplify the many terms containing abcabc.

Show solutionHide solution

Solution

Idea. The constraint kills most terms after expansion.

Clearing denominators. Let D=∏(a2+a+1)D=\prod(a^2+a+1). The inequality is
∑cyc(b2+b+1)(c2+c+1)−D≥0.\sum_{\rm cyc}(b^2+b+1)(c^2+c+1)-D\ge0.

Expansion. Expanding mechanically gives a polynomial whose terms, after substituting abc=1abc=1, simplify:
- −a2b2c2=−1-a^2b^2c^2=-1,
- −a2b2c−a2bc2−ab2c2=−ab−ac−bc-a^2b^2c-a^2bc^2-ab^2c^2=-ab-ac-bc,
- −a2bc−ab2c−abc2=−a−b−c-a^2bc-ab^2c-abc^2=-a-b-c,
- −abc=−1-abc=-1.

Combining with the positive terms a2+b2+c2+a+b+c+2a^2+b^2+c^2+a+b+c+2:
a2+b2+c2−ab−bc−ca ≥0.a^2+b^2+c^2-ab-bc-ca\ \ge0.

Equality. Iff a=b=c=1a=b=c=1.

Lesson. With a multiplicative constraint, expand fully and use abc=1abc=1 to cancel: often the inequality collapses to a classical one such as a2+b2+c2≥ab+bc+caa^2+b^2+c^2\ge ab+bc+ca.