SOLUTION: If a, b, and c are consecutive positive integers, show that 6/5<=A/(B+C) +B/(C+A)+C/(A+B)<=2

Question 34649: If a, b, and c are consecutive positive integers, show that 6/5<=A/(B+C) +B/(C+A)+C/(A+B)<=2
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
6/5<=A/(B+C) +B/(C+A)+C/(A+B)<=2
LET S=A/(B+C) +B/(C+A)+C/(A+B)={A/(B+C)+1}+{B/(C+A)+1}+{C/(A+B)+1}-3
S=(A+B+C){1/(B+C)+1/(C+A)+1/(A+B)}-3
WE HAVE B=A+1,C=A+2..SO
S=(A+A+1+A+2){1/(2A+3)+1/(2A+2)+1/(2A+1)}-3
S=3(A+1){1/(2A+3)+1/(2A+2)+1/(2A+1)}-3
SINCE A,B,C ARE POSITIVE INTEGERS MIIMUM VALUE OF A IS 1 AND MAXIMUM
IS TENDING TO INFINITY�..
IF WE PUT L=1/(2A+3)+1/(2A+2)+1/(2A+1),THEN WE FIND 1/(2A+1) IS THE
BIGGEST OF THE 3 FRACTIONS AND 1/(2A+3) IS THE SMALLEST OF THE 3
FRACTIONS.HENCE
L-MAX =3/(2A+1)�..AND�.. L-MIN=3/(2A+3)
S-MAX=9(A+1)/(2A+1)-3�.AND�.S-MIN=9(A+1)/(2A+3)-3
IF WE TEST THESE AT A=1 AND A TENDING TO INFINITY WE FIND THAT
S-MAX=9*2/3-3=3�OR�TOWARDS INFINITY�S-MAX=9/2-3=3/2=1.5
S-MIN =9*2/5-3=3/5�.OR..TOWARDS INFINITY�S-MIN=9/2-3=3/2
HENCE WE CAN CONCLUDE THAT S-MAX IS 3/2 IN THE LEAST OR 3 AT THE HIGHEST.
AND S- MIN IS NEVER LESSTHAN 3/5
OR 3/5<=S<=3/2
SO OBVIOUSLY 6/5<=S<=2
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