Simple trigonometric substitutions with broad results

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1 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 tan β + tan β tan γ + tan γ tan α =. Proof. First of all note that if α = β = γ, then the statement clearly holds. Assume without loss of generality that α β. Because 0 < α + β < π, 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 ( π x), we have tan γ = cot α + β = tan α tan β tan α + tan β, yielding Now suppose that tan α tan β + tan β tan γ + tan γ tan α =. () tan α tan β + tan β tan γ + tan γ tan α =, () 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 triangle. Subtracting () from () we get tan γ γ = tan. Thus γ γ = kπ for some nonnegative integer k. But γ γ γ + γ < π, so it follows that k = 0. That is γ = γ, as desired. Mathematical Reflections 6 (007)

2 Theorem. Let α, β, γ be angles in (0, π). Then α, β, γ are the angles of a triangle if and only if sin α + sin β + sin γ + sin α sin β sin γ =. Proof. As 0 < α + β < π, 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 sin γ + sin α sin β γ sin = cos α + β ( cos α + β + sin α sin β ) = cos α + β cos α β cos α + cos β = ( ) ( ) sin α + sin β = = sin α sin β. Thus Now suppose that sin α + sin β γ + sin + sin α sin β γ sin =. () sin α + sin β + sin γ + sin α sin β sin γ =, () for some α, β, γ in (0, π). Subtracting () from () we obtain sin γ γ sin + sin α sin β ) γ γ (sin sin = 0, that is ( sin γ ) γ γ γ sin (sin + sin + sin α sin β ) = 0. The second factor can be written as sin γ γ + sin + cos α β cos α + β = sin γ + cos α β, which is clearly greater than 0. It follows that sin γ that α, β, γ are the angles of a triangle. = sin γ, and so γ = γ, showing Mathematical Reflections 6 (007)

3 Substitutions and Transformations T. Let α, β, γ be angles of a triangle. Let A = π α, B = π β, C = π γ. Then A + B + C = π, and 0 A, B, C < π. This transformation allows us to switch from angles of an arbitrary triangle to angles of an acute triangle. Note that cyc(sin α = cos A), cyc(cos α = sin A), cyc(tan α = cot A), cyc(cot α = tan A), where by cyc we denote a cyclic permutation of angles. T. 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. S. Let a, b, c be positive real numbers such that ab + bc + ca =. Using the function f : (0, π ) (0, + ), for f(x) = tan x, we can do the following substitution a = tan α, b = tan β, c = tan γ, where α, β, γ are the angles of a triangle ABC. S. Let a, b, c be positive real numbers such that ab + bc + ca =. Applying T to S, 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 abc it follows that bc + ca + ab =. Due to S, we can substitute that is where α, β, γ are the angles of a triangle. a = tan α, b = tan β, c = tan γ, a = cot α, b = cot β, c = cot γ, S4. Let a, b, c be positive real numbers such that a + b + c = abc. Applying T to S3, we have a = tan A, b = tan B, c = tan C, Mathematical Reflections 6 (007) 3

4 where A, B, C are the angles of an acute triangle. S5. Let a, b, c be positive real numbers such that a + b + c + abc =. Note that since all the numbers are positive it follows that a, b, c <. Usign the function f : (0, π) (0, ), for f(x) = sin x, and recalling Theorem, we can substitute where α, β, γ are the angles of a triangle. a = sin α, b = sin β, c = sin γ, S6. Let a, b, c be positive real numbers such that a +b +c +abc =. Applying T 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 T to expressions yz (x + y)(x + z), zx (y + z)(y + x), xy (z + x)(z + y), they can be substituted by (s b)(s c), bc (s c)(s a), ca (s a)(s b), ab where a, b, c are the sidelengths of a triangle. Recall the following identities sin α = (s b)(s c) bc Thus our expressions can be substituted by where α, β, γ are the angles of a triangle. sin α, sin β, sin γ,, cos α s(s a) =. bc 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 cos α, cos β, cos γ, where α, β, γ are the angles of a triangle. Mathematical Reflections 6 (007) 4

5 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. cos α + cos β + cos γ sin α + sin β + sin γ 3. sin α + sin β + sin γ cos α + cos β + cos γ cos α cos β cos γ sin α sin β sin γ 8 4. sin α sin β sin γ cos α cos β cos γ cot α + cot β + cot C cos α + cos β + cos γ sin α + sin β + sin C sin α + sin β + sin γ cos α + cos β + cos γ cot α + cot β + cot γ tan α + tan β + tan γ 3 Well-known identities Let α, β, γ be angles of a triangle ABC. Then. cos α + cos β + cos γ = + 4 sin α sin β sin γ. sin α + sin β + sin γ = 4 cos α cos β cos γ 3. sin α + sin β + sin γ = 4 sin α sin β sin γ 4. sin α + sin β + sin γ = + cos α cos β cos γ For arbitrary angles α, β, γ we have sin α + sin β + sin γ sin(α + β + γ) = 4 sin α + β cos α + cos β + cos γ + cos(α + β + γ) = 4 cos α + β sin β + γ cos β + γ sin γ + α. cos γ + α. Mathematical Reflections 6 (007) 5

6 Applications. Let x, y, z be positive real numbers. Prove that x x + (x + y)(x + z) + y y + (y + z)(y + x) + Solution. The inequality is equivalent to (x + y)(x + z) x = z x + (z + x)(z + y). (Walther Janous, Crux Mathematicorum). (x+y)(x+z) + x Beceause the inequality is homogeneous, we can assume that xy +yz +zx =. Let us apply substitution S: cyc(x = tan α ), where α, β, γ are angles of a triangle. We get ( tan α + tan β ) ( tan α + tan γ ) tan α = sin α, and similar expressions for the other terms. The inequality becomes sin α + sin α + sin β + sin β + sin γ + sin γ, that is + sin α + + sin β + + sin γ. On the other hand, using the well-known inequality sin α + sin β + sin γ 3 and the Cauchy-Schwarz inequality, we have 9 ( + sin α ) + ( + sin β ) + ( + sin γ ) + sin α, and we are done.. Let x, y, z be real numbers greater than such that x + y + z =. Prove that x + y + z x + y + z. (Iran, 997) Mathematical Reflections 6 (007) 6

7 Solution. Let (x, y, z) = (a +, b +, c + ), with a, b, c positive real numbers. Note that the hypothesis is equivalent to ab + bc + ca + abc =. Then it suffices to prove that a + b + c a + b + c + 3. Squaring both sides of the inequality and canceling some terms yields ab + bc + ca 3. Using substitution S5 we get (ab, bc, ca) = (sin α, sin β, sin γ ), where ABC is an arbitrary triangle. The problem reduces to proving that sin α + sin β + sin γ 3, which is well-known and can be done using Jensen inequality. 3. Let a, b, c be positive real numbers such that a + b + c =. Prove that ab bc ca c + ab + a + bc + b + ca 3. (Open Olympiad of FML No-39, Russia) Solution. The inequality is equivalent to ab ((c + a)(c + b) + bc (a + b)(a + c) + ca (b + c)(b + a) 3. Substitution S7 replaces the three terms in the inequality by sin α, sin β, sin γ. Thus it suffices to prove sin α + sin β + sin γ 3, which clearly holds. 4. Let a, b, c be positive real numbers such that (a + b)(b + c)(c + a) =. Prove that ab + bc + ca 3 4. (Cezar Lupu, Romania, 005) Solution. Observe that the inequality is equivalent to ( ab ) 3 ( 3 4 ) 3 (a + b) (b + c) (c + a). Mathematical Reflections 6 (007) 7

8 Because the inequality is homogeneous, we can assume that ab + bc + ca =. We use substitution S: cyc(a = tan α ), where α, β, γ are the angles of a triangle. Note that (a + b)(b + c)(c + a) = ( ) cos γ cos α cos β = cos α cos β cos γ. Thus it suffices to prove that ( ) or cos α cos β cos γ 4 cos α cos β cos γ 3 3. From the identity 4 cos α cos β cos γ = sin α + sin β + sin γ, the inequality is equivalent to sin α + sin β + sin γ 3 3. But f(x) = sin x is a concave function on (0, π) and the conclusion follows from Jensen s inequality. 5. Let a, b, c be positive real numbers such that a + b + c =. Prove that a + b + c + 3abc., (Poland, 999) ( ) bc Solution. Let cyc x = a. It follows that cyc(a = yz). The inequality becomes x y + y z + x z + 3xyz, where x, y, z are positive real numbers such that xy + yz + zx =. Note that the inequality is equivalent to (xy + yz + zx) + 3xyz + xyz(x + y + z), or 3 x + y + z. Applying substitution S cyc(x = tan α ), it suffices to prove tan α + tan β + tan γ 3. The last inequality clearly holds, as f(x) = tan x is convex function on (0, π), and the conclusion follows from Jensen s inequality. Mathematical Reflections 6 (007) 8

9 6. 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) + Applying substitution S8, it suffices to prove that cos α + cos β + cos γ, z(x + y + z) (z + x)(z + y). where α, β, γ are the angles of a triangle. Using transformation T 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 α π sin α α for all α (0, π ). The conclusion immediately follows. 7. Let a, b, c be positive real numbers such that a + b + c + = 4abc. Prove that a + b + c ab bc ca (Daniel Campos Salas, Mathematical Reflections, 007) Solution. Rewrite the condition as bc + ca + ab + abc = 4. Observe that we can use substitution S5 in the following way ( bc, ca, ) ( = sin α ab, β sin, γ ) sin, Mathematical Reflections 6 (007) 9

10 where α, β, γ are angles of a triangle. It follows that ( a, b, ) ( sin β = sin γ c sin α, sin γ sin α sin β, sin β sin ) α sin γ. Then it suffices to prove that sin β sin γ sin α + sin γ sin α sin β + sin β sin α sin γ 3 sin α + sin β + sin γ. The right-hand side of the inequality is well known. For the left-hand side we use trasnformation T 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 x y + z + y x + z + z x + y 3, which a famous Nesbitt s inequality, and we are done. 8. Let a, b, c (0, ) be real numbers such that ab + bc + ca =. Prove that a a + b b + c c 3 ( a + b + ) c. 4 a b c (Calin Popa) Solution. We apply substitution S cyc(a tan A ), where A, B, C are angles of a triangle. Because a, b, c (0, ), it follows that tan A, tan B, tan C (0, ), that is A, B, C are angles of an acute triangle. Note that ( a cyc a = sin A cos ) A = tan A. cos A Thus the inequality is equivalent to tan A + tan B + tan C 3 ( tan A + tan B + ). tan C Now observe that if we apply transformation T and the result in Theorem, 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) 3 (tan A tan B + tan B tan C + tan A tan C). This can be written as (tan A tan B) + (tan B tan C) + (tan C tan A) 0, and we are done. Mathematical Reflections 6 (007) 0

11 9. Let x, y, z be positive real numbers. Prove that y + z z + x x + y 6(x + y + z) x y z 3(x + y)(y + z)(z + x). (Vo Quoc Ba Can, Mathematical Reflections, 007) Solution. Note that the inequality is equivalent to (x + y)(z + x) 4(x + y + z) (y + z). x(x + y + z) 3 cyc Let use transfromation T and substitution S8. We get ( ) (x + y)(z + x) cyc (y + z) x(x + y + z) = a cos α = 4R sin α, and 4(x + y + z) 4R(sin α + sin β + sin γ) =, 3 3 where α, β, γ are angles of a triangle with circumradius R. Therefore it suffices to prove that ( 3 sin α + sin β + sin γ ) sin α cos α + sin β cos β + sin γ cos γ. Because f(x) = cos x is a concave function on [0, π], from Jensen s inequality we obtain 3 ( cos α 3 + cos β + cos γ ). Finally, we observe that f(x) = sin x is an increasing function on [0, π], while g(x) = cos x is a decreasing function on [0, π]. Using Chebyschev s inequality, we have ( sin α 3 + sin β + sin γ ) ( cos α + cos β + cos γ ) sin α cos α, and the conclusion follows. Mathematical Reflections 6 (007)

12 Problems for independent study. Let a, b, c be positive real numbers such that a + b + c =. Prove that a b c b + c c + a a + b. (Romanian Mathematical Olympiad, 005). Let a, b, c be positive real numbers such that a + b + c =. Prove that a b + b c + c a 6 (A. Teplinsky, Ukraine, 005) 3. Let a, b, c be positive real numbers such that ab + bc + ca =. Prove that a + b b + c c + a 4. Prove that for all positive real numbers a, b, c, (a + )(b + )(c + ) 9(ab + bc + ca). (Le Trung Kien) (APMO, 004) 5. Let x, y, z be positive real numbers such that x + y + z =. Prove that x + yz + x + yz + x + yz xyz + x + y + z. 6. Let a, b, c be positive real numbers. Prove that b + c + c + a + a + b ( a 4 a b c b + c + b c + a + c ). a + b (APMO, 00) (Mircea Lascu) 7. Let a, b, c be positive real numbers, such that a + b + c = abc. Prove that ab + bc + ca 9(a + b + c). (Belarus, 996) Mathematical Reflections 6 (007)

13 8. Let a, b, c be positive real numbers. Prove that b + c + c + a + a + b abc + a b c (a + b)(b + c)(c + a) (Bui Viet Anh) 9. Let a, b, c be positive real numbers such that a + b + c = abc. Prove that (a )(b )(c ) (Gabriel Dospinescu, Marian Tetiva) 0. Let a, b, c be nonnegative real numbers such that a + b + c + abc = 4. Prove that 0 ab + bc + ca abc. (Titu Andreescu, USAMO, 00) REFERENCES [. ] Titu Andreescu, Zuming Feng, 03 Trigonometry Problems: From the Training of the USA IMO Team, Birkhauser, 004 [. ] Titu Andreescu, Vasile Cirtoaje, Gabriel Dospinescu, Mircea Lascu - Old and New Inequalities, GIL Publishing House, 004 [3. ] Tran Phuong, Diamonds in Mathematical Inequalities, Hanoi Publishing House, 007 [4. ] N.M. Sedrakyan Geometricheskie Neravenstva, Yerevan, 004 [5. ] E. Specht, Collected Inequalities, [6. ] H. Lee, Topics in Inequalities - Theorems and Techniques [7. ] H. Lee, Inequalities through problems [8. ] Crux Mathematicorum and Mathematical Mayhem (Canada) Mathematical Reflections 6 (007) 3

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