Advanced book on Mathematics Olympiad

Real Analysis 569

513.Fixα, β, γand consider the function

f (x, y, z)=

cosx

sinα

+

cosy

sinβ

+

cosz

sinγ

with the constraintsx+y+z=π,x, y, z≥0. We want to determine the maximum of
f (x, y, z). In the interior of the triangle described by the constraints a maximum satisfies

sinx

sinα

=−λ,

siny

sinβ

=−λ,

sinz

sinβ

=−λ,

x+y+z=π.

By the law of sines, the triangle with anglesx, y, zis similar to that with anglesα, β, γ,
hencex=α,y=β, andz=γ.
Let us now examine the boundary. Ifx=0, then cosz=−cosy. We prove that

1

sinα

+cosy

(

1

sinβ

−

1

sinγ

)

<cotα+cotβ+cotγ.

This is a linear function in cosy, so the inequality will follow from the corresponding
inequalities at the two endpoints of the interval, namely from

1

sinα

+

1

sinβ

−

1

sinγ

<cotα+cotβ+cotγ

and

1

sinα

+

1

sinβ

−

1

sinγ

<cotα+cotβ+cotγ.

By symmetry, it suffices to prove just one of these two, the first for example. Eliminating
the denominators, we obtain

sinβsinγ+sinαsinγ−sinαsinβ<sinβsinγcosα+sinαsinγcosβ

+sinαsinβcosγ.

The laws of sine and cosine allow us to transform this into the equivalent

bc+ac−ab <

b^2 +c^2 −a^2

2

+

a^2 +c^2 −b^2

2

+

a^2 +b^2 −c^2

2

,