Expert: Paul Klarreich - 6/22/2009

Question
QUESTION: given: a^2 + b^2 = 1; c^2 + d^2 = 1

prove ac + bd is less than  or equal to 1

ANSWER: Questioner:   James
Country:  United States
Category:  Advanced Math
Private:  No
 
Subject:  inequal;ities proof
Question:  given: a^2 + b^2 = 1; c^2 + d^2 = 1

prove ac + bd is less than  or equal to 1
....................................................
Hi, James,

If a^2 + b^2 = 1, then certainly |a| <= 1.  

So write  a = sin t, for some t.  Then b = cos t.

Likewise,  write  c = sin s,  d = cos s.

So ac + bd =  sin t sin s + cos t cos s = cos(s - t) <= 1.

Cute problem.


---------- FOLLOW-UP ----------

QUESTION: can the problem be solver without the substitution?
I have gotten this far.

a^2 - b^2 =/< 1; c^2 - d^2 =/< 1; a^2,b^2,d^2,c^2 =/<1

(a^2 - b^2)(c^2 - d^2) =/<  1

expand

a^2c^2 - b^2c^2 - d^2a^2 + b^2d^2 =/<1

a^2c^2 - b^2d^2 =/< 1 + b^2d^2 + d^2a^2

a^2c^2 + B^2d^2 =/<1 because the other terms are positive

a^2c^2 +2A62b^2d^2c^2 + B62d62 =/< 1 + 2 a^2b^2d^2c^2

(ac + bd)^2 =/< 1 because the other teerms are positive

Here is my problem ;

ac+bd=/<1 or ac + bd =/>1 how do I exclude this alternative?

Answer
Hi, James,

Sorry I don't have a way to work out your proof, but here is another proof based on two things:

1. Ptolemy's Theorem, which you can look up.
2. The Law of cosines.

See the attached picture (if I remember to attach it)
............................................
Ptolemy's Theorem: (see basic diagram)

If a quadrilateral is inscribed in a circle then the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals.

In the above, AB CD + AD BC = AC BD
.........................................
In this picture, the opposite sides are  x opp 1,  and a opp d, while the diagonals are b,c.  Since a^2 + b^2 = 1, and c^2 + d^2 = 1, the triangles ABD and ACD are inscribed in the semicircle whose diameter is 1.

P's Thm gives:

bc = 1x + ad

or  x = bc - ad.

So call angle BDC  Angle t and use the Law of Cosines to write:

x^2 = b^2 + d^2 - 2bd cos t  (I)

Now if  x = bc - ad, then square that to get:

x^2 = b^2 c^2 + a^2 d^2 - 2abcd

Substitute into (I):

b^2 c^2 + a^2 d^2 - 2abcd = b^2 + d^2 - 2bd cos t

2bd cos t - 2abcd = b^2(1 - c^2) + d^2(1 - a^2)  << Bring terms over.

Remember that 1 - c^2 = d^2, and that  1 - a^2 = b^2:

2bd cos t - 2abcd = b^2(d^2) + d^2(b^2)

2bd cos t - 2abcd = 2b^2d^2  << combine

2bd (cos t - ac) = 2b^2d^2   << factor on the left


cos t - ac = bd   << cancel  2bd

cos t = ac + bd   << bring over ac

So ac + bd  = cos t <= 1

(I still like my first proof better.)