Prove Triangle of Sides with Length...

Date: 8/1/96 at 21:27:38
From: Anonymous
Subject: Prove Triangle of Sides with Length...

I have a problem for you from the Asian Pacific Mathematics Olympiad
held on March 12, 1996:

Question 5:

Let a, b and c be the lengths of the sides of a triangle. Prove that
the square root of (a+b-c) plus the square root of (b+c-a) plus the
square root of (c+a-b) is equal to or less than the sum of the square
roots of a, b and c. Determine when equality occurs.

Intuitively, equality occurs when the triangle is equilateral. My 
teacher has spent weeks on it but has been unable to reach a final 
proof.

Would you please please please please please please please please
please please please please please please please please please please
please please pretty please help me to solve the problem.

Date: 8/2/96 at 11:25:18
From: Doctor Anthony
Subject: Re: Prove Triangle of Sides with Length...

The information that a, b, c are the sides of a triangle has no 
significance beyond the fact that all the terms like a+b-c will be 
positive (two sides of a triangle are together greater than the third 
side).  So we let

a+b-c = p^2
b+c-a = q^2
c+a-b = r^2    From these equations  2a = r^2+p^2
                                     2b = p^2+q^2
                                     2c = q^2+r^2

So a = (r^2+p^2)/2    b = (p^2+q^2)/2     c = (q^2+r^2)/2  and we are 
asked to prove that:

     sqrt(a) + sqrt(b) + sqrt(c) >or= p + q + r  ...(1)

Now use the fact that (x^2+y^2)/2 > {(x+y)/2}^2  (See later for proof 
of this)
 
Substituting for a, b, c in expression(1) we have:

 sqrt{(r^2+p^2)/2} + sqrt{(p^2+q^2)/2} + sqrt{(q^2+r^2)/2}

          >or= sqrt{(r+p)/2}^2 + sqrt{(p+q)/2}^2 + sqrt{(q+r)/2}^2

          >or= (1/2){r+p + p+q + q+r}

          >or= (1/2){2p + 2q + 2r}

          >or= {p + q + r}                       


And this is the required inequality, i.e.

sqrt(a) + sqrt(b) + sqrt(c) >or= sqrt(a+b-c) + sqrt(b+c-a) + 
sqrt(c+a-b)


NOTE:  We must show that (x^2+y^2)/2 > {(x+y)/2}^2  as promised above.

This inequality requires  2(x^2+y^2) > x^2 + 2xy + y^2

                         x^2 - 2xy + y^2 > 0

                          (x-y)^2 > 0    and this is always true.  
Hence the inequality is true.


-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/