Advanced book on Mathematics Olympiad

36 2 Algebra

111. Prove that
     sin^3 a
     sinb

+

cos^3 a

cosb

≥sec(a−b),

for alla, b∈( 0 ,π 2 ).

112. Prove that
     1
     a+b

+

1

b+c

+

1

c+a

+

1

23

√

abc

≥

(a+b+c+^3

√

abc)^2

(a+b)(b+c)(c+a)

,

for alla, b, c >0.

2.1.4 The Triangle Inequality

In its most general form, the triangle inequality states that in a metric spaceXthe distance

functionδsatisfies

δ(x, y)≤δ(x, z)+δ(z, y), for anyx, y, z∈X.

An equivalent form is

|δ(x, y)−δ(y, z)|≤δ(x, z).

Here are some familiar examples of distance functions: the distance between two real

or complex numbers as the absolute value of their difference, the distance between two

vectors inn-dimensional Euclidean space as the length of their difference‖v−w‖, the

distance between two matrices as the norm of their difference, the distance between two

continuous functions on the same interval as the supremum of the absolute value of their

difference. In all these cases the triangle inequality holds.

Let us see how the triangle inequality can be used to solve a problem from

T.B. Soulami’s bookLes olympiades de mathématiques: Réflexes et stratégies(Ellipses,

1999).

Example.For positive numbersa, b, cprove the inequality

√

a^2 −ab+b^2 +

√

b^2 −bc+c^2 ≥

√

a^2 +ac+c^2.

Solution.The inequality suggests the following geometric construction. With the same

originO, draw segmentsOA,OB, andOCof lengthsa, b, respectively,c, such that

OBmakes 60◦angles withOAandOC(see Figure 12).

The law of cosines in the trianglesOAB,OBC, andOACgivesAB^2 =a^2 −ab+b^2 ,

BC^2 =b^2 −bc+c^2 , andAC^2 =a^2 +ac+c^2. Plugging these formulas into the triangle

inequalityAB+BC≥ACproduces the inequality from the statement.