Advanced book on Mathematics Olympiad

Geometry and Trigonometry 661

674.By eventually changingφ(t)toφ(t)+θ 2 , whereθis the argument of 4P^2 − 2 Q,
we may assume that 4P^2 − 2 Qis real and positive. We can then ignore the imaginary
parts and write

4 P^2 − 2 Q= 4

(∫∞

0

e−tcosφ(t)dt

) 2

− 4

(∫∞

0

e−tsinφ(t)dt

) 2

− 2

∫∞

0

e−^2 tcos 2φ(t)dt.

Ignore the second term. Increase the first term using the Cauchy–Schwarz inequality:

(∫∞

0

e−tcosφ(t)dt

) 2

=

(∫∞

0

e−

(^12) t
e−
(^12) t
cosφ(t)dt

) 2

≤

(∫∞

0

e−tdt

)(∫∞

0

e−tcos^2 φ(t)dt

)

=

∫∞

0

e−tcos^2 φ(t)dt.

We then have

4 P^2 − 2 Q≤ 4

∫∞

0

e−tcos^2 φ(t)dt− 2

∫∞

0

e−^2 tcos 2φ(t)dt

= 4

∫∞

0

(e−t−e−^2 t)cos^2 φ(t)dt+ 1

≤ 4

∫∞

0

(e−t−e−^2 t)dt+ 1 = 3.

Equality holds only when cos^2 φ(t)=1 for allt, and in general ifφ(t)is constant.
(K. Löwner, from G. Pólya, G. Szego, ̋ Aufgaben und Lehrsätze aus der Analysis,
Springer-Verlag, 1964)

675.The given inequality follows from the easier
√
ab+

√

( 1 −a)( 1 −b)≤ 1.

To prove this one, leta=sin^2 αandb=sin^2 β,α, β∈[ 0 ,π 2 ]. The inequality becomes
sinαsinβ+cosαcosβ≤1, or cos(α−β)≤1, and this is clearly true.

676.First, note that ifx>2, thenx^3 − 3 x> 4 x− 3 x=x>

√

x+2, so all solutionsx
should satisfy− 2 ≤x≤2. Therefore, we can substitutex=2 cosafor somea∈[ 0 ,π].
Then the given equation becomes

2 cos 3a=

√

2 ( 1 +cosa)=2 cos

a

2

,