HCA Mathematics 3: Inequalities (2025 syllabus)

Notes for the Singapore-Cambridge ‘A’-Level Inequalities 9820 Mathematics III (2025 onward) Gerard Sayson <geruls@broskiclan.org> Syllabus requirements More inequalities outside the syllabus will be included, and some of the in- equalities below will be detailed at greater depth. ˆ AM-GM inequality ˆ Cauchy-Schwarz inequality ˆ Triangle inequality Read this first before continuing This document is only meant to cover inequalities in the H3 syllabus. It may not be enough to cover inequalities expected at mathematical olympiads, although this can be a great starting point. Most non-trivial inequalities here are proven. If you would like to see resources on mathematical olympiad inequali- ties, the author recommends Yufei Zhao’s handout on inequalities and Evan Chen’s brief notes. 1

Contents 1 The AM-GM inequality and other extensions 3 1.1 Proof of the AM-GM inequality . . . . . . . . . . . . . . . . . 3 1.2 Weighted AM-GM inequality . . . . . . . . . . . . . . . . . . 5 1.3 QM-AM-GM-HM inequality chain . . . . . . . . . . . . . . . 5 2 The Cauchy-Schwarz inequality 7 2.1 Proof of the inequality . . . . . . . . . . . . . . . . . . . . . . 7 2.2 Titu’s lemma and Nesbitt’s inequality . . . . . . . . . . . . . 9 3 The triangle inequality 10 3.1 Two-variable triangle inequality . . . . . . . . . . . . . . . . . 10 3.2 Generalized triangle inequality . . . . . . . . . . . . . . . . . 11 2

1 The AM-GM inequality and other extensions The arithmetic mean-geometric meaninequality is one of the most important inequalities in mathematics – whether it be in Olympiads or in general, it is used widely in proofs involving non-negative real numbers. For starters, we have the following definitions: Definition 1.1. The arithmetic mean of a1, a2, . . . , an ∈ R+ 0 is defined as AM = a1 + a2 + · · ·+ an n = 1 n nX k=1 ak Definition 1.2. The geometric mean of a1, a2, . . . , an ∈ R+ 0 is defined as GM = n√a1a2 . . . an = n vuut nY k=1 ak 1.1 Proof of the AM-GM inequality Now, we will prove the AM-GM inequality. Consider the simple case n = 2, where a1 + a2 2 ≥ √a1a2 This holds if and only if a2 1 + 2a1a2 + a2 2 4 ≥ a1a2 a2 1 + 2a1a2 + a2 2 − 4a1a2 ≥ 0 a2 1 − 2a1a2 + a2 2 ≥ 0 (a1 − a2)2 ≥ 0 which is obviously true since a1, a2 ∈ R. Theorem 1.3 (AM-GM inequality). For anya1, a2, . . . , an ∈ R+ 0 , a1 + a2 + · · ·+ an n ≥ n√a1a2 . . . an with equality attained if and only ifa1 = a2 = · · ·= an. Proof. The above inequality can be reduced to the inequality a1 + a2 + · · ·+ an n n ≥ a1a2 . . . an on which we proceed by induction. 3

The base case n = 1 results in equality. Hence, suppose that the in- equality holds for some arbitrary n. Then, consider the case n + 1, with a1, a2, . . . , an, an+1 ∈ R+ 0 . We know that α = a1 + a2 + · · ·+ an + an+1 n + 1 is their mean. If a1 = a2 = · · ·= an = an+1, then there is equality. Otherwise, there exists ai < α < aj. Since addition and multiplication are commutative, the list a1, . . . , an, an+1 can be reordered such that ai =