wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Triangle inequality
        
(Message started by: NickH on Apr 23rd, 2003, 11:41am)
Title: Triangle inequality
Post by NickH on Apr 23rd, 2003, 11:41am
The sides of a triangle have lengths a, b, c.  Show that:

3/2 <= a/(b + c) + b/(c + a) + c/(a + b) < 2.
Title: Re: Triangle inequality
Post by Mond on Apr 23rd, 2003, 9:45pm

This should work... the greater than applies to any 3 positive reals, but the lesser than inequality requires that they're sides of a triangle.
     
[hide]
the greater than case :
    a/(b+c) = (a+b+c)/(b+c) - 1 ....
 
  this gives

      (a+b+c) (1/(b+c) + 1/(c+a) + 1/(a+b)) - 3
   
  =0.5((a+b)+(b+c)+(c+a)) (1/(b+c) + 1/(c+a) + 1/(a+b)) -3
 
  since AM >= HM , (x+y+z)(1/x+1/y+1/z) >= 9
 
  so the above is   >= 9/2 - 3  (= 3/2)
 
   the equality is when the triangle is equilateral

the lesser than :
       
     since in a triangle,  a+b > (a+b+c)/2  (because a+b>c)
 
          a/(b+c) < a/s , where s = (a+b+c)/2
     
   so the sum  < (a+b+c)/s  (=2)  [/hide]
                   
               
   
       


Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board