Simple trigonometric substitutions with broad results Vardan Verdiyan, Daniel Campos Salas Often, the key to solve some intricate algebraic inequality is to simplify it by employing a trigonometric substitution When we make a clever trigonometric substitution the problem may reduce so much that we can see a direct solution immediately Besides, trigonometric functions have well-known properties that may help in solving such inequalities As a result, many algebraic problems can be solved by using an inspired substitution We start by introducing the readers to such substitutions After that we present some well-known trigonometric identities and inequalities Finally, we discuss some Olympiad problems and leave others for the reader to solve Theorem Let α, β, γ be angles in (0, π) Then α, β, γ are the angles of a triangle if and only if tan α β β γ γ α tan + tan tan + tan tan = 2 2 2 Proof First of all note that if α = β = γ, then the statement clearly holds Assume without loss of generality that α = β Because < α + β < 2π, it follows that there exists an angle in (−π, π), say γ , such that α + β + γ = π Using the addition formulas and the fact that tan x = cot π2 − x , we have α β − tan tan γ α+β 2, tan = cot = α β 2 tan + tan 2 yielding tan β β γ γ α α tan + tan tan + tan tan = 2 2 2 (1) tan α β β γ γ α tan + tan tan + tan tan = 1, 2 2 2 (2) Now suppose that for some α, β, γ in (0, π) We will prove that γ = γ , and this will imply that α, β, γ are the angles of a γ−γ γ γ triangle Subtracting (1) from (2) we get tan = tan Thus = kπ for 2 γ−γ γ γ some nonnegative integer k But ≤ + < π, so it follows that k = 2 That is γ = γ , as desired Mathematical Reflections (2007) Theorem Let α, β, γ be angles in (0, π) Then α, β, γ are the angles of a triangle if and only if sin2 α β γ α β γ + sin2 + sin2 + sin sin sin = 2 2 2 Proof As < α + β < 2π, there exists an angle in (−π, π), say γ , such that α + β + γ = π Using the product-to-sum and the double angle formulas we get sin2 γ α β γ + sin sin sin 2 2 α+β α+β α β cos + sin sin 2 2 α+β α−β = cos cos 2 cos α + cos β = = cos − sin2 = α + − sin2 β 2 α β = − sin2 − sin2 2 Thus sin2 α β γ α β γ + sin2 + sin2 + sin sin sin = 2 2 2 (1) β γ α β γ α + sin2 + sin2 + sin sin sin = 1, 2 2 2 (2) Now suppose that sin2 for some α, β, γ in (0, π) Subtracting (1) from (2) we obtain sin2 that is sin γ α β γ − sin2 + sin sin 2 2 γ γ − sin 2 sin sin γ γ − sin 2 γ γ α β + sin + sin sin 2 2 = 0, = The second factor can be written as sin γ γ α−β α+β γ α−β + sin + cos − cos = sin + cos , 2 2 2 which is clearly greater than It follows that sin γ2 = sin γ2 , and so γ = γ , showing that α, β, γ are the angles of a triangle Mathematical Reflections (2007) Substitutions and Transformations T1 Let α, β, γ be angles of a triangle Let A= π−α π−β π−γ , B= , C= 2 Then A + B + C = π, and ≤ A, B, C < π2 This transformation allows us to switch from angles of an arbitrary triangle to angles of an acute triangle Note that cyc(sin α α α α = cos A), cyc(cos = sin A), cyc(tan = cot A), cyc(cot = tan A), 2 2 where by cyc we denote a cyclic permutation of angles T2 Let x, y, z be positive real numbers Then there is a triangle with sidelengths a = x + y, b = y + z, c = z + x This transformation is sometimes called Dual Principle Clearly, s = x+y+z and (x, y, z) = (s−a, s−b, s−c) This transformation already triangle inequality S1 Let a, b, c be positive real numbers such that ab + bc + ca = Using the function f : (0, π2 ) → (0, +∞), for f (x) = tan x, we can the following substitution a = tan α , b = tan β , γ c = tan , where α, β, γ are the angles of a triangle ABC S2 Let a, b, c be positive real numbers such that ab + bc + ca = Applying T1 to S1, we have a = cot A, b = cot B, c = cot C, where A, B, C are the angles of an acute triangle S3 Let a, b, c be positive real numbers such that a + b + c = abc Dividing by 1 abc it follows that bc + ca + ab = Due to S1, we can substitute α = tan , a that is β = tan , b α β , b = cot , 2 where α, β, γ are the angles of a triangle a = cot γ = tan , c γ c = cot , S4 Let a, b, c be positive real numbers such that a + b + c = abc Applying T1 to S3, we have a = tan A, b = tan B, c = tan C, Mathematical Reflections (2007) where A, B, C are the angles of an acute triangle S5 Let a, b, c be positive real numbers such that a2 + b2 + c2 + 2abc = Note that since all the numbers are positive it follows that a, b, c < Usign the function f : (0, π) → (0, 1), for f (x) = sin x2 , and recalling Theorem 2, we can substitute a = sin α , b = sin β , γ c = sin , where α, β, γ are the angles of a triangle S6 Let a, b, c be positive real numbers such that a2 +b2 +c2 +2abc = Applying T1 to S5, we have a = cos A, b = cos B, c = cos C, where A, B, C are the angles of an acute triangle S7 Let x, y, z be positive real numbers Applying T2 to expressions yz , (x + y)(x + z) zx , (y + z)(y + x) xy , (z + x)(z + y) (s − c)(s − a) , ca (s − a)(s − b) , ab they can be substituted by (s − b)(s − c) , bc where a, b, c are the sidelengths of a triangle Recall the following identities sin α = (s − b)(s − c) α , cos = bc s(s − a) bc Thus our expressions can be substituted by sin α β γ , sin , sin , 2 where α, β, γ are the angles of a triangle S8 Analogously to S7, the expressions x(x + y + z) , (x + y)(x + z) y(x + y + z) , (y + z)(y + x) z(x + y + z) , (z + x)(z + y) can be substituted by α β γ , cos , cos , 2 where α, β, γ are the angles of a triangle cos Mathematical Reflections (2007) Further we present a list of inequalities and equalities that can be helpful in solving many problems or simplify them Well-known inequalities Let α, β, γ be angles of a triangle ABC Then α β γ cos α + cos β + cos γ ≤ sin + sin + sin ≤ 2 2 √ α β γ 3 sin α + sin β + sin γ ≤ cos + cos + cos ≤ 2 2 α β γ cos α cos β cos γ ≤ sin sin sin ≤ 2 √ α β γ 3 sin α sin β sin γ ≤ cos cos cos ≤ 2 √ α β C cot + cot + cot ≥ 3 2 α β C cos2 α + cos2 β + cos2 γ ≥ sin2 + sin2 + sin2 ≥ 2 α β γ sin2 α + sin2 β + sin2 γ ≤ cos2 + cos2 + cos2 ≤ 2 α β γ √ cot α + cot β + cot γ ≥ tan + tan + tan ≥ 2 Well-known identities Let α, β, γ be angles of a triangle ABC Then cos α + cos β + cos γ = + sin sin α + sin β + sin γ = cos α α cos sin β β cos sin γ γ sin 2α + sin 2β + sin 2γ = sin α sin β sin γ sin2 α + sin2 β + sin2 γ = + cos α cos β cos γ For arbitrary angles α, β, γ we have α+β β+γ γ+α sin sin 2 α+β β+γ γ+α cos α + cos β + cos γ + cos(α + β + γ) = cos cos cos 2 sin α + sin β + sin γ − sin(α + β + γ) = sin Mathematical Reflections (2007) Applications Let x, y, z be positive real numbers Prove that x x+ (x + y)(x + z) + y y+ (y + z)(y + x) + z x+ (z + x)(z + y) ≤ (Walther Janous, Crux Mathematicorum) Solution The inequality is equivalent to 1+ ≤ (x+y)(x+z) x2 Beceause the inequality is homogeneous, we can assume that xy + yz + zx = Let us apply substitution S1: cyc(x = tan α2 ), where α, β, γ are angles of a triangle We get (x + y)(x + z) = x2 tan α β + tan 2 tan2 α γ + tan 2 tan = α sin2 α, and similar expressions for the other terms The inequality becomes sin α2 sin β2 sin γ2 + + ≤ 1, + sin α2 + sin γ2 + sin β2 that is 2≤ 1 + + β + sin α2 + sin γ2 + sin On the other hand, using the well-known inequality sin α2 + sin β2 + sin γ2 ≤ and the Cauchy-Schwarz inequality, we have 2≤ (1 + sin α 2) + (1 + sin β 2) + (1 + sin γ 2) ≤ , + sin α2 and we are done Let x, y, z be real numbers greater than such that √ x−1+ y−1+ √ z−1≤ √ x + y1 + z1 = Prove that x + y + z (Iran, 1997) Mathematical Reflections (2007) Solution Let (x, y, z) = (a + 1, b + 1, c + 1), with a, b, c positive real numbers Note that the hypothesis is equivalent to ab + bc + ca + 2abc = Then it suffices to prove that √ √ √ √ a + b + c ≤ a + b + c + Squaring both sides of the inequality and canceling some terms yields √ √ ab + bc + √ ca ≤ Using substitution S5 we get (ab, bc, ca) = (sin2 α2 , sin2 β2 , sin2 γ2 ), where ABC is an arbitrary triangle The problem reduces to proving that sin α β γ + sin + sin ≤ , 2 2 which is well-known and can be done using Jensen inequality Let a, b, c be positive real numbers such that a + b + c = Prove that ab + c + ab bc + a + bc ca ≤ b + ca (Open Olympiad of FML No-239, Russia) Solution The inequality is equivalent to ab + ((c + a)(c + b) bc + (a + b)(a + c) ca ≤ (b + c)(b + a) Substitution S7 replaces the three terms in the inequality by sin α2 , sin β2 , sin γ2 Thus it suffices to prove sin α2 + sin β2 + sin γ2 ≤ 23 , which clearly holds Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) = Prove that ab + bc + ca ≤ (Cezar Lupu, Romania, 2005) Solution Observe that the inequality is equivalent to ab Mathematical Reflections (2007) ≤ (a + b)2 (b + c)2 (c + a)2 Because the inequality is homogeneous, we can assume that ab + bc + ca = We use substitution S1: cyc(a = tan α2 ), where α, β, γ are the angles of a triangle Note that cos γ2 (a + b)(b + c)(c + a) = cos α cos β = cos α cos β2 cos γ2 Thus it suffices to prove that or ≤ cos2 α2 cos2 β cos2 γ , √ α β γ 3 cos cos cos ≤ 2 2 From the identity cos α2 cos β2 cos γ2 = sin α + sin β + sin γ, the inequality is equivalent to √ 3 sin α + sin β + sin γ ≤ But f (x) = sin x is a concave function on (0, π) and the conclusion follows from Jensen’s inequality Let a, b, c be positive real numbers such that a + b + c = Prove that √ a2 + b2 + c2 + 3abc ≤ (Poland, 1999) Solution Let cyc x = becomes bc a It follows that cyc(a = yz) The inequality √ x2 y + y z + x2 z + 3xyz ≤ 1, where x, y, z are positive real numbers such that xy + yz + zx = Note that the inequality is equivalent to √ (xy + yz + zx)2 + 3xyz ≤ + 2xyz(x + y + z), or √ ≤ x + y + z Applying substitution S1 cyc(x = tan α2 ), it suffices to prove tan α β γ √ + tan + tan ≥ 2 The last inequality clearly holds, as f (x) = tan x2 is convex function on (0, π), and the conclusion follows from Jensen’s inequality Mathematical Reflections (2007) Let x, y, z be positive real numbers Prove that x(y + z) + y(z + x) + z(x + y) ≥ (x + y)(y + z)(z + x) x+y+z (Darij Grinberg) Solution Rewrite the inequality as x(x + y + z) + (x + y)(x + z) y(x + y + z) + (y + z)(y + x) z(x + y + z) ≥ (z + x)(z + y) Applying substitution S8, it suffices to prove that cos β γ α + cos + cos ≥ 2, 2 where α, β, γ are the angles of a triangle Using transformation T1 cyc(A = π−α ), where A, B, C are angles of an acute triangle, the inequality is equivalent to sin A + sin B + sin C ≥ There are many ways to prove this fact We prefer to use Jordan’s inequality, that is 2α π ≤ sin α ≤ α for all α ∈ (0, ) π The conclusion immediately follows Let a, b, c be positive real numbers such that a + b + c + = 4abc Prove that 1 1 1 + + ≥3≥ √ +√ +√ a b c ca ab bc (Daniel Campos Salas, Mathematical Reflections, 2007) Solution Rewrite the condition as 1 1 + + + = bc ca ab abc Observe that we can use substitution S5 in the following way 1 , , bc ca ab Mathematical Reflections (2007) = sin2 α β γ , sin2 , sin2 2 , where α, β, γ are angles of a triangle It follows that 1 , , a b c = sin β2 sin γ2 sin γ2 sin α2 sin β2 sin α2 , , sin α2 sin γ2 sin β Then it suffices to prove that sin γ2 sin α2 sin β2 sin α2 sin β2 sin γ2 α β γ + + ≥ ≥ sin + sin + sin γ α β sin sin 2 2 sin The right-hand side of the inequality is well known For the left-hand side we use trasnformation T2 backwards Denote by x = s − a, y = s − b, z = s − c, where s is the semiperimeter of the triangle The left-hand side is equivalent to y z x + + ≥ , y+z x+z x+y which a famous Nesbitt’s inequality, and we are done Let a, b, c ∈ (0, 1) be real numbers such that ab + bc + ca = Prove that a b c + + ≥ 2 1−a 1−b 1−c − a2 − b2 − c2 + + a b c (Calin Popa) Solution We apply substitution S1 cyc(a ≡ tan A2 ), where A, B, C are angles of a triangle Because a, b, c ∈ (0, 1), it follows that tan A2 , tan B2 , tan C2 ∈ (0, 1), that is A, B, C are angles of an acute triangle Note that cyc sin A2 cos A2 a tan A = = − a2 cos A Thus the inequality is equivalent to tan A + tan B + tan C ≥ 1 + + tan A tan B tan C Now observe that if we apply transformation T1 and the result in Theorem 1, we get tan A + tan B + tan C = tan A tan B tan C Hence our inequality is equivalent to (tan A + tan B + tan C)2 ≥ (tan A tan B + tan B tan C + tan A tan C) This can be written as (tan A − tan B)2 + (tan B − tan C)2 + (tan C − tan A)2 ≥ 0, and we are done Mathematical Reflections (2007) 10 Let x, y, z be positive real numbers Prove that y+z + x z+x + y x+y ≥ z 16(x + y + z)3 3(x + y)(y + z)(z + x) (Vo Quoc Ba Can, Mathematical Reflections, 2007) Solution Note that the inequality is equivalent to (x + y)(z + x) 4(x + y + z) √ ≥ x(x + y + z) (y + z) cyc Let use transfromation T2 and substitution S8 We get cyc (y + z) and (x + y)(z + x) a α = α = 4R sin x(x + y + z) cos 2 , 4(x + y + z) 4R(sin α + sin β + sin γ) √ √ = , 3 where α, β, γ are angles of a triangle with circumradius R Therefore it suffices to prove that √ α β γ α α β β γ γ sin + sin + sin ≥ sin cos + sin cos + sin cos 2 2 2 2 2 Because f (x) = cos x2 is a concave function on [0, π], from Jensen’s inequality we obtain √ α β γ ≥ cos + cos + cos 2 Finally, we observe that f (x) = sin x2 is an increasing function on [0, π], while g(x) = cos x2 is a decreasing function on [0, π] Using Chebyschev’s inequality, we have sin α β γ + sin + sin 2 cos α β γ + cos + cos 2 ≥ sin α α cos , 2 and the conclusion follows Mathematical Reflections (2007) 11 Problems for independent study Let a, b, c be positive real numbers such that a + b + c = Prove that √ c b a +√ +√ ≥ c + a b+c a+b (Romanian Mathematical Olympiad, 2005) Let a, b, c be positive real numbers such that a + b + c = Prove that −1 a −1+ b −1 b −1+ c −1 c −1≥6 a (A Teplinsky, Ukraine, 2005) Let a, b, c be positive real numbers such that ab + bc + ca = Prove that √ 1 1 +√ +√ ≥2+ √ c+a a+b b+c (Le Trung Kien) Prove that for all positive real numbers a, b, c, (a2 + 2)(b2 + 2)(c2 + 2) ≥ 9(ab + bc + ca) (APMO, 2004) Let x, y, z be positive real numbers such that x1 + y1 + z1 = Prove that √ √ √ √ √ √ √ x + yz + x + yz + x + yz ≥ xyz + x + y + z (APMO, 2002) Let a, b, c be positive real numbers Prove that b+c c+a a+b + + ≥4 a b c a b c + + b+c c+a a+b (Mircea Lascu) Let a, b, c be positive real numbers, such that a + b + c = √ abc Prove that ab + bc + ca ≥ 9(a + b + c) (Belarus, 1996) Mathematical Reflections (2007) 12 Let a, b, c be positive real numbers Prove that b+c c+a a+b + + +2 a b c abc ≥2 (a + b)(b + c)(c + a) (Bui Viet Anh) Let a, b, c be positive real numbers such that a + b + c = abc Prove that √ (a − 1)(b − 1)(c − 1) ≤ − 10 (Gabriel Dospinescu, Marian Tetiva) 10 Let a, b, c be nonnegative real numbers such that a2 + b2 + c2 + abc = Prove that ≤ ab + bc + ca − abc ≤ (Titu Andreescu, USAMO, 2001) REFERENCES [1 ] Titu Andreescu, Zuming Feng, 103 Trigonometry Problems: From the Training of the USA IMO Team, Birkhauser, 2004 [2 ] Titu Andreescu, Vasile Cirtoaje, Gabriel Dospinescu, Mircea Lascu - Old and New Inequalities, GIL Publishing House, 2004 [3 ] Tran Phuong, Diamonds in Mathematical Inequalities, Hanoi Publishing House, 2007 [4 ] N.M Sedrakyan “Geometricheskie Neravenstva”, Yerevan, 2004 [5 ] E Specht, “Collected Inequalities”, http://www.imo.org.yu/othercomp/Journ/ineq.pdf [6 ] H Lee, “Topics in Inequalities - Theorems and Techniques” [7 ] H Lee, “Inequalities through problems” [8 ] Crux Mathematicorum and Mathematical Mayhem (Canada) Mathematical Reflections (2007) 13