Triangle Proofs

Date: 06/28/98 at 04:03:32
From: Dr. H. M. Behl
Subject: Problem with a question

Dear Dr. Math,

Could you please help me with these two questions? I cannot figure out 
how to start doing the problems.

1. If the sides of a triangle are a, b, c, then prove that
	
	(a+b+c)^3 >= 27(b+c-a)(c+a-b)(a+b-c)

2. If a>0, b>0, c>0 then prove that

	a/b+c  +  b/c+a  +  c/a+b   >= 3/2


Thanking you and waiting for your reply,

Rohit Behl

Date: 06/28/98 at 08:57:36
From: Doctor Anthony
Subject: Re: Problem with a question

>(1) If the sides of a triangle are a, b, c, then prove that
	 (a+b+c)^3 >= 27(b+c-a)(c+a-b)(a+b-c)

Use the fact that the arithmetic mean of three positive numbers is 
greater than the geometric mean.

So    x+y+z
     -------  > (xyz)^(1/3)
        3

     (x+y+z)^3 > 27(xyz)   ........ (1)

Now let  x = a+b-c   y = b+c-a    z = c+a-b

(since a,b,c are sides of a triangle, and two sides are always greater 
than the third side,  x, y, z are positive)

     x+y+z = 2a+2b+2c-a-b-c = a+b+c  and we get, substituting into (1)

    (a+b+c)^3 > 27(a+b-c)(b+c-a)(c+a-b) 


>2. If a>0, b>0, c>0 then prove that
	a/b+c  +  b/c+a  +  c/a+b   >= 3/2

To prove this we need first to show that the least value of the 
lefthand side occurs when  a/(b+c) = b/(c+a) = c/(a+b).

If a+b+c = constant and if we assume that c is fixed, then a+b is 
fixed but we vary a and b to minimize the lefthand side, so
c/(a+b) = constant and we consider the sum of the other two fractions.

      a/(b+c) + b/(c+a)

       a(c+a) + b(b+c)      a^2 + b^2 + c(a+b)
    =  --------------   =   -------------------   
         (b+c)(c+a)            (b+c)(c+a)

                         (a+b)^2 - 2ab + c(a+b)     
                     =   -----------------------    remember
                              (b+c)(c+a)            a+b is fixed,

the top line is a minimum if 2ab is a maximum, and the bottom line is 
a maximum if (b+c)(c+a) is a maximum, i.e. if the two factors are 
equal. Both these conditions require a = b. With a = b, if we now 
allow c to vary, a similar argument leads to c = b, and so for the 
least value on the lefthand side we require a = b = c.

Then with a = b = c we obtain the value of the lefthand side:

     1/2 + 1/2 + 1/2  =  3/2    and this is the minimum value.

It follows that  a/(b+c) + b/(c+a) + c/(a+b)  >= 3/2

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/