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Corollary 1.2.1. Let p, q, and r be positive real numbers. Let θ 1 , θ 2 , and θ 3 be real numbers satisfying θ 1 + θ 2 + θ 3 = π. Then, the following inequality holds. p cos θ 1 + q cos θ 2 + r cos θ 3 ≤ 1 ( qr 2 p + rp q + pq ) . r (√ √ Proof. Take (x, y, z) = qr p , rp q , √ ) pq r and apply the above proposition. Theorem 1.2.2. (Barrow’s Inequality) Let P be an interior point of a triangle ABC and let U, V , W be the points where the bisectors of angles BP C, CP A, AP B cut the sides BC,CA,AB respectively. Prove that P A + P B + P C ≥ 2(P U + P V + P W ). Proof. ([MB] and [AK]) Let d 1 = P A, d 2 = P B, d 3 = P C, l 1 = P U, l 2 = P V , l 3 = P W , 2θ 1 = ∠BP C, 2θ 2 = ∠CP A, and 2θ 3 = ∠AP B. We need to show that d 1 + d 2 + d 3 ≥ 2(l 1 + l 2 + l 3 ). It’s easy to deduce the following identities l 1 = 2d 2d 3 d 2 + d 3 cos θ 1 , l 2 = 2d 3d 1 d 3 + d 1 cos θ 2 , and l 3 = 2d 1d 2 d 1 + d 2 cos θ 3 , By the AM-GM inequality and the above corollary, this means that l 1 + l 2 + l 3 ≤ √ d 2 d 3 cos θ 1 + √ d 3 d 1 cos θ 2 + √ d 1 d 2 cos θ 3 ≤ 1 2 (d 1 + d 2 + d 3 ) . As another application of the above trigonometric proposition, we establish the following inequality Corollary 1.2.2. ([AK], Abi-Khuzam) Let x 1 , · · · , x 4 be positive real numbers. Let θ 1 , · · · , θ 4 be real numbers such that θ 1 + · · · + θ 4 = π. Then, √ (x 1 x 2 + x 3 x 4 )(x 1 x 3 + x 2 x 4 )(x 1 x 4 + x 2 x 3 ) x 1 cos θ 1 + x 2 cos θ 2 + x 3 cos θ 3 + x 4 cos θ 4 ≤ . x 1 x 2 x 3 x 4 √ Proof. Let p = x12 2 +x 2 2x 1 x 2 + x32 2 +x 4 2x 3 x 4 q = x1x2+x3x4 2 and λ = p q . In the view of θ 1 + θ 2 + (θ 3 + θ 4 ) = π and θ 3 + θ 4 + (θ 1 + θ 2 ) = π, the proposition implies that x 1 cos θ 1 + x 2 cos θ 2 + λ cos(θ 3 + θ 4 ) ≤ pλ = √ pq, and x 3 cos θ 3 + x 4 cos θ 4 + λ cos(θ 1 + θ 2 ) ≤ q λ = √ pq. Since cos(θ 3 + θ 4 ) + cos(θ 1 + θ 2 ) = 0, adding these two above inequalities yields √ x 1 cos θ 1 + x 2 cos θ 2 + x 3 cos θ 3 + x 4 cos θ 4 ≤ 2 √ (x 1 x 2 + x 3 x 4 )(x 1 x 3 + x 2 x 4 )(x 1 x 4 + x 2 x 3 ) pq = . x 1 x 2 x 3 x 4 10
1.3 Applications of Complex Numbers In this section, we discuss some applications of complex numbers to geometric inequality. Every complex number corresponds to a unique point in the complex plane. The standard symbol for the set of all complex numbers is C, and we also refer to the complex plane as C. The main tool is applications of the following fundamental inequality. Theorem 1.3.1. If z 1 , · · · , z n ∈ C, then |z 1 | + · · · + |z n | ≥ |z 1 + · · · + z n |. Proof. Use induction on n with the triangle inequality. Theorem 1.3.2. (Ptolemy’s Inequality) For any points A, B, C, D in the plane, we have AB · CD + BC · DA ≥ AC · BD. Proof. Let a, b, c and 0 be complex numbers that correspond to A, B, C, D in the complex plane. It becomes |a − b| · |c| + |b − c| · |a| ≥ |a − c| · |b|. Applying the Triangle Inequality to the identity (a − b)c + (b − c)a = (a − c)b, we get the result. Problem 6. ([TD]) Let P be an arbitrary point in the plane of a triangle ABC with the centroid G. Show the following inequalities (1) BC · P B · P C + AB · P A · P B + CA · P C · P A ≥ BC · CA · AB and (2) P A 3 · BC + P B 3 · CA + P C 3 · AB ≥ 3P G · BC · CA · AB. Solution. We only check the first inequality. Regard A, B, C, P as complex numbers and assume that P corresponds to 0. We’re required to prove that |(B − C)BC| + |(A − B)AB| + |(C − A)CA| ≥ |(B − C)(C − A)(A − B)|. It remains to apply the Triangle Inequality to the identity (B − C)BC + (A − B)AB + (C − A)CA = −(B − C)(C − A)(A − B). Problem 7. (IMO Short-list 2002) Let ABC be a triangle for which there exists an interior point F such that ∠AF B = ∠BF C = ∠CF A. Let the lines BF and CF meet the sides AC and AB at D and E, respectively. Prove that AB + AC ≥ 4DE. Solution. Let AF = x, BF = y, CF = z and let ω = cos 2π 3 + i sin 2π 3 . We can toss the pictures on C so that the points F , A, B, C, D, and E are represented by the complex numbers 0, x, yω, zω 2 , d, and e. It’s an easy exercise to establish that DF = xz x+z and EF = xy xz xy x+y . This means that d = − x+z ω and e = − x+y ω. We’re now required to prove that |x − yω| + |zω 2 − x| ≥ 4 −zx ∣z + x ω + xy ∣ ∣∣∣ x + y ω2 . Since |ω| = 1 and ω 3 = 1, we have |zω 2 − x| = |ω(zω 2 − x)| = |z − xω|. Therefore, we need to prove |x − yω| + |z − xω| ≥ 4zx ∣z + x − 4xy ∣ ∣∣∣ x + y ω . ∣ More strongly, we establish that |(x − yω) + (z − xω)| ≥ ∣ 4zx z+x − 4xy ∣∣ x+y ω or |p − qω| ≥ |r − sω|, where p = z + x, q = y + x, r = 4zx 4xy z+x and s = x+y . It’s clear that p ≥ r > 0 and q ≥ s > 0. It follows that |p − qω| 2 − |r − sω| 2 = (p − qω)(p − qω) − (r − sω)(r − sω) = (p 2 − r 2 ) + (pq − rs) + (q 2 − s 2 ) ≥ 0. It’s easy to check that the equality holds if and only if △ABC is equilateral. 11
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Page 3 and 4: Chapter 1 Geometric Inequalities It
Page 5 and 6: Exercise 3. (Darij Grinberg) Let a,
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Page 9 and 10: 1.2 Trigonometric Methods In this s
Page 11: Alternatively, one may obtain anoth
Page 15 and 16: which follows from the AM-GM inequa
Page 17 and 18: 2.2 Algebraic Substitutions We know
Page 19 and 20: Proof 3. As in the previous proof,
Page 21 and 22: Problem 16. (IMO Short-list 2001) L
Page 23 and 24: (IMO 2000/2) Let a, b, c be positiv
Page 25 and 26: Proof. We don’t employ the Ravi s
Page 27 and 28: Third Solution. (by an IMO 2005 con
Page 29 and 30: Remark 3.1.1. When does the equalit
Page 31 and 32: 3.2 Schur’s Inequality and Muirhe
Page 33 and 34: Problem 23. (Tournament of Towns 19
Page 35 and 36: Second Solution. It’s equivalent
Page 37 and 38: Second Solution. After setting a =
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Page 41 and 42: (Iran 1998) Prove that, for all x,
Page 43 and 44: Let a, b > 0 and m, n ∈ N. Take x
Page 45 and 46: (3) ⇒ (2) : Let x 1 , · · · ,
Page 47 and 48: 4.2 Power Means Convexity is one of
Page 49 and 50: 4.3 Majorization Inequality We say
Page 51 and 52: Lemma 4.4.1. Let f : (a, b) −→
Page 53 and 54: M 6. (Korea 2001) Let x 1 , · ·
Page 55 and 56: M 24. (IMO Short List 1998) Let r 1
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Page 59 and 60: M 57. (China 2002) Given c ∈ ( 1
Page 61 and 62: P 12. Putnam 98B1 Find the minimum
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P 33. (MM Q197, Norman Schaumberger
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