Advanced book on Mathematics Olympiad

Geometry and Trigonometry 663

=sinusin(v+w)−cosucos(v+w)=−cos(u+v+w).

And of course this takes values in the interval[− 1 , 1 ]. The inequality is proved.
(T. Andreescu, Z. Feng, 103Trigonometry Problems, Birkhäuser 2004)
680.The denominators suggest the substitution based on tangents. This idea is further
enforced by the identityx+y+z=xyz, which characterizes the tangents of the angles
of a triangle. Setx=tanA,y=tanB,z=tanC, withA, B, Cthe angles of an acute
triangle. Note that
tanA
√
1 +tan^2 A

=

tanA

secA

=sinA,

so the inequality is equivalent to

sinA+sinB+sinC≤

3

√

3

2

.

This is Jensen’s inequality applied to the functionf(x)=sinx, which is concave

on( 0 ,π 2 ).

681.If we multiply the inequality through by 2, thus obtaining

2 x

1 −x^2

+

2 y

1 −y^2

+

2 z

1 −z^2

≥ 3

√

3 ,

the substitution by tangents becomes transparent. This is because we should recognize

the double-angle formulas on the left-hand side.

The conditions 0< x,y,z <1 andxy+xz+yz=1 characterize the tangents of

the half-angles of an acute triangle. Indeed, ifx=tanA 2 ,y=tanB 2 , andz=tanC 2 , then

0 < x,y,z <1 impliesA, B, C∈( 0 ,π 2 ). Also, the equalityxy+xz+yz=1, which

is the same as

1

z

=

x+y

1 −xy

,

implies

cot

C

2

=tan

A+B

2

.

And this is equivalent toπ 2 −C 2 =A+ 2 B,orA+B+C=π.
Returning to the problem, with the chosen trigonometric substitution the inequality
assumes the much simpler form

tanA+tanB+tanC≥ 3

√

3.