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Problems > The Maine Page > 2003

Fractional System

Solve the system of equations in positive real a,b,c:

a

b

+

b

a

+

a

c

+

c

a

+

b

c

+

c

b

= 6

a

bc

+

b

ac

+

c

ab

= 4

Please note: A complete solution must demonstrate that you have all possible solutions!

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Fractional System Solution

First, we make note of the fact that:

a

b

+

b

a

>= 2to prove this: assume

a

b

+

b

a

(a² + b²)

ab

a² + b²a² + b² - 2ab (a - b)²Similarly,

a

c

+

c

a

>=2, and

b

c

+

c

b

>= 2

Since the first equation is the sum of three quantities which are all greater than or equal to two, and their sum is 6, we conclude that each of these quantities must be equal to two.

Thus

a

b

+

b

a

= 2

(a² + b²)

ab

= 2
a² + b² = 2ab
a² + b² - 2ab = 0
(a - b)² = 0
a = bSimilar reasoning gives us b = c

Now we simply need to substutite a everywhere in the second equation:

a

a²

+

a

a²

+

a

a²

= 4

3

a

= 4
a =

3

4

Solution: (
3
4
,
3
4
,
3
4
)

Logs Within Logs

A Recursive Partitioned Function

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