560 Real Analysis
∂f
∂A(A, B)=−sinA+sin(A+B)= 0 ,
∂f
∂B(A, B)=−sinB+sin(A+B)= 0.From here we obtain sinA=sinB=sin(A+B), which can happen only ifA=B=π 3.
This is the unique critical point, for whichf(π 3 ,π 3 )=^32. On the boundary, ifA=0or
B=0, thenf(A,B)=1. Same ifA+B=π. We conclude that the maximum of
cosA+cosB+cosCis^32 , attained for the equilateral triangle, while the minimum is 1,
which is attained only for a degenerate triangle in which two vertices coincide.
505.We rewrite the inequality as
sinαcosβcosγ+cosαsinβcosγ+cosαcosβsinγ≤2
√
3
,
and prove it forα, β, γ∈[ 0 ,π 2 ]. To this end, we denote the left-hand side byf (α, β, γ )
and find its maximum in the specified region. The critical points in the interior of the
domain are solutions to the system of equations
cosαcosβcosγ−sinαsinβcosγ−sinαcosβsinγ= 0 ,
−sinαsinβcosγ+cosαcosβcosγ−cosαsinβsinγ= 0 ,
−sinαcosβsinγ−cosαsinβsinγ+cosαcosβcosγ= 0.Bring this system into the form
cosαcosβcosγ=sinαsin(β+γ),
cosαcosβcosγ=sinβsin(γ+α),
cosαcosβcosγ=sinγsin(α+β).From the first two equations, we obtain
sinα
sin(α+γ)=
sinβ
sin(β+γ).
The functiong:( 0 ,π 2 ),g(t)=sinsin(t+tγ)is strictly increasing, since
g′(t)=costsin(t+γ)−sintcos(t+γ)
(sin(t+γ))^2=
sinγ
(sin(t+γ))^2> 0.
Henceg(α)=g(β)impliesα=β. Similarly,β=γ. The condition that(α, α, α)is
a critical point is the trigonometric equation cos^3 α=sinαsin 2α, which translates into
cos^3 α= 2 ( 1 −cos^2 α)cosα. We obtain cosα=
√
2
3 , andf (α, α, α)=
√^2
3. This will