Advanced book on Mathematics Olympiad

Real Analysis 561

be the maximum once we check that no value on the boundary of the domain exceeds
this number.
But when one of the three numbers, sayα, is zero, thenf( 0 ,β,γ)=sin(β+γ)≤1.
Also, ifα=π 2 , thenf(π 2 ,β,γ)=cosβcosγ≤1. Hence the maximum offis√^23 and
the inequality is proved.
506.Ifabcd=0 the inequality is easy to prove. Indeed, if sayd =0, the inequality
becomes 3(a^2 −ab+b^2 )c^2 ≥ 2 a^2 c^2 , which is equivalent to the obvious

c^2

((

a−

3

2

b

) 2

+

3

4

b^2

)

≥ 0.

Ifabcd  =0, divide through byb^2 d^2 and setx=ab,y=dc. The inequality becomes

3 (x^2 −x+ 1 )(y^2 −y+ 1 )≥ 2 ((xy)^2 −xy+ 1 ),

or

3 (x^2 −x+ 1 )(y^2 −y+ 1 )− 2 ((xy)^2 −xy+ 1 )≥ 0.

The expression on the left is a two-variable functionf (x, y)defined on the whole plane.

To find its minimum we need to determine the critical points. These are solutions to the

system of equations

∂f

∂x

(x, y)= 2 (y^2 − 3 y+ 3 )x−( 3 y^2 − 5 y+ 3 )= 0 ,

∂f

∂y

(x, y)= 2 (x^2 − 3 x+ 3 )y−( 3 x^2 − 5 x+ 3 )= 0.

The system can be rewritten as

2 x=

3 y^2 − 5 y+ 3

y^2 − 3 y+ 3

,

2 y=

3 x^2 − 5 x+ 3

x^2 − 3 x+ 3

,

or

2 x= 3 +

4 y− 6

y^2 − 3 y+ 3

,

2 y= 3 +

4 x− 6

x^2 − 3 x+ 3

.

A substitution becomes natural:u= 2 x−3,v= 2 y−3. The system now reads