Question 34660: If a,b, and c are consective positive integers, show that (6/5)<= (a/(b+c)) + (b/(a+c)) + (c/(a+b)) <= 2
Answer by mukhopadhyay(490)   (Show Source):
You can put this solution on YOUR website! a,b, and c are consective positive integers
If a = x-1 (x-1 positive integer), b = x, and c = x+1;
(a/(b+c)) + (b/(a+c)) + (c/(a+b))
= (x-1)/(2x+1) + x/(2x) + (x+1)/(2x-1)
= (x-1)/(2x+1) + 1/2 + (x+1)/(2x-1)
= 1/2 + (x-1)/(2x+1) + (x+1)/(2x-1)
= 1/2 + [(x-1)(2x-1)+(x+1)(2x+1)]/[(2x-1)(2x+1)]
= 1/2 + [(2x^2-2x-x+1)+(2x^2+2x+x+1)]/[(2x-1)(2x+1)]
= 1/2 + [4x^2+2]/[4x^2-1]......... (Exp 0.5)
= 1/2 + [4x^2-1+3]/[4x^2-1]
= 1/2 + 1 + 3/[4x^2-1]
= 3/2 + 3/[4x^2-1].....(Exp 1)
......................
x-1 is a positive integer (>= 1)
=> x >= 2
=> x^2 >= 4
=> 4x^2 > 16
=> 4x^2 - 1 >= 15
=> 1/[4x^2 - 1] <= 1/15
=> (Exp 1) < 3/2 + 1/15 (which is <= 2)
The above proves the upper boundary of the expression in question.
..........................
Same way, 3/2 + 3/[4x^2-1] >= 3/2 (because 3/[4x^2-1] >=0)
3/2 >= 6/5
This proves the lower boundary of the expression.
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