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 = an < αand aj = an+1 > αwithout loss of generality. Since an+1 − α >0 and an − α > 0, it follows that ( an+1 − α)(an − α) > 0. We know that α(n + 1) = a1 + · · ·+ an+1 so α = a1 + · · ·+ an−1 + (an + an+1 − α) n . Due to the induction hypothesis, one has αn+1 = αnα ≥ a1a2 . . . an−1(an + an+1 − α)α. We use the result (an+1 − α)(an − α) > 0 to justify that (an+1 − α)(an − α) = α(an + an+1 − α) − anan+1 > 0 implying ( an + an+1 − α)α > anan+1. Hence αn+1 = αnα ≥ a1a2 . . . an−1(an + an+1 − α)α ≥ a1a2 . . . anan+1 and the proof is complete. Example 1.4 (IMO 2020 Shortlisted Problem) . Suppose a, b, c, d∈ R+ satisfy ac+bd = (a+c)(b+d). Find the smallest possible value of a b + b c + c d + d a . Solution. We know that a b + b c + c d + d a = a b + c d

+ b c + d a

and a b + c d

+ b c + d a

≥ 2 r ac bd + r bd ac ! Clearly, r ac bd + r bd ac = √ac√ bd + √ bd√ac = √ac√ bd + √ bd√ac = ac + bd√ abcd = (a + c)(b + d)√ abcd Using again the AM-GM inequality, a + c ≥ 2√ac and b + d ≥ 2 √ bd so a b + b c + c d + d a ≥ 2 (a + c)(b + d)√ abcd

≥ 2 (2√ac)(2 √ bd)√ abcd ! = 8 and so the minimum value is 8. The AM-GM inequality appears in the formula booklet, unlike the more powerful inequalities like the weighted AM-GM inequality and the QM-AM- GM-HM inequality chain. 4

1.2 Weighted AM-GM inequality Theorem 1.3 implies the weighted AM-GM inequality, which states: Corollary 1.5 (Weighted AM-GM inequality). For anyw1, w2, . . . , wn ≥ 0 corresponding toa1, a2, . . . , an ≥ 0, w1a1 + w2a2 + · · ·+ wnan Σw ≥ Σw q aw1 1 aw2 2 . . . awn n where Σw = w1 + · · ·+ wn, with equality attained only whena1 = a2 = · · ·= an. A proof will not be provided here. 1.3 QM-AM-GM-HM inequality chain The quadratic mean-arithmetic mean-geometric mean-harmonic meanin- equality chain, also known as the mean inequality chain, state the relation- ship between the harmonic mean, geometric mean, arithmetic mean, and quadratic mean of positive real numbers. This time, no number can be zero, or the harmonic mean would be undefined. Definition 1.6. The quadratic mean (root mean square) of a1, a2, . . . , an ∈ R+ is defined as QM = r a2 1 + a2 2 + · · ·+ a2n n = vuut 1 n nX k=1 a2 k Definition 1.7. The harmonic mean of a1, a2, . . . , an ∈ R+ is defined as HM = n 1 a1 + 1 a2 + · · ·+ 1 an In order to prove that QM ≥ AM ≥ GM ≥ HM, one must prove sepa- rately that QM ≥ AM, then AM ≥ GM implies QM ≥ AM ≥ GM. Finally, proving GM ≥ HM completes the inequality chain. Lemma 1.8 (QM-AM). For anya1, a2, . . . , an ∈ R+, r a2 1 + a2 2 + · · ·+ a2n n ≥ a1 + a2 + · · ·+ an n Proof. This reduces to the inequality a2 1 + a2 2 + · · ·+ a2 n n ≥ a2 1 + a2 2 + · · ·+ a2 n n2 ≥ a1 + a2 + · · ·+ an n 2 5

which holds due to n ≥ 1, n∈ Z+ and Theorem 2.1, which asserts nX k=0 ukvk !2 ≤ nX k=0 u2 k ! nX k=0 v2 k ! for uk, vk ∈ R. Lemma 1.9 (GM-HM). For anya1, a2, . . . , an ∈ R+, n√a1a2 . . . an ≥ n 1 a1 + 1 a2 + · · ·+ 1 an Proof. This inequality reduces to 1 n√a1a2 . . . an ≤ 1 a1 + 1 a2 + · · ·+ 1 an n However, 1 n√a1a2 . . . an = ( n√a1a2 . . . an)−1 = (a1a2 . . . an)− 1 n = r 1 a1 1 a2 . . .1 an and thus the inequality above holds, due to the AM-GM inequality. Theorem 1.10 (QM-AM-GM-HM). For anya1, a2, . . . , an ∈ R+, r a2 1 + a2 2 + · · ·+ a2n n ≥ a1 + a2 + · · ·+ an n ≥ n√a1a2 . . . an ≥ n 1 a1 + 1 a2 + · · ·+ 1 an where no number is zero. Proof. This follows by Lemma 1.8 and Lemma 1.9, as well as Theorem 1.3. 6

2 The Cauchy-Schwarz inequality The Cauchy-Schwarz inequality (or Cauchy-Bunyakovsky-Schwarz inequal- ity) provides an upper bound on the inner product between two vectors in an inner product space in terms of the product of the vector norms (or magnitudes). Formally, an inner product spaceis a real or complex vector space (per- mitting vector arithmetic) bundled with an operation called the inner prod- uct. An inner product is usually denoted ⟨⃗ u, ⃗ v⟩. The inner product of Rn satisfies the following conditions: ˆ ⟨⃗ u, ⃗ v⟩ = ⟨⃗ v, ⃗ u⟩. In other words, the inner product is commutative for real vectors ⃗ u, ⃗ v. ˆ ⟨a⃗ x+ b⃗ y, ⃗ z⟩ = a⟨⃗ x, ⃗ z⟩ + b⟨⃗ y, ⃗ z⟩ for scalars a, b∈ R. ˆ ⟨⃗ x, ⃗ x⟩ > 0 if ⃗ xis not the zero vector. We will operate in the real vector space Rn where vectors are of length n, and we use the dot product as the inner product; all the properties above are satisfied. 2.1 Proof of the inequality Theorem 2.1 (Cauchy-Schwarz inRn). For anyu1, . . . , un and v1, . . . , vn ∈ R, nX k=0 ukvk !2 ≤ nX k=0 u2 k ! nX k=0 v2 k ! with equality if and only if there existsc ̸= 0 such that uk = cvk for all i = 1, 2, . . . , n. Proof. Consider the real vectors ⃗ a=   u1 u2 ... un   ⃗b =   v1 v2 ... vn   of length n, where ui, vi ∈ R for any i = 1, 2, . . . , n. Then, if |⃗ x| denotes the Euclidean norm |⃗ x| = p x2 1 + · · ·+ x2n, |⃗ a||⃗b| cos(θ) = ⃗ a· ⃗b = u1v1 + u2v2 + · · ·+ unvn and so q u2 1 + u2 2 + · · ·+ u2n q v2 1 + v2 2 + · · ·+ v2n cos(θ) = u1v1 + u2v2 + · · ·+ unvn 7

where θ is the angle between ⃗ aand ⃗b