document.write( "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 \n" ); document.write( "
Algebra.Com's Answer #20921 by venugopalramana(3286)\"\" \"About 
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6/5<=A/(B+C) +B/(C+A)+C/(A+B)<=2
\n" ); document.write( "LET S=A/(B+C) +B/(C+A)+C/(A+B)={A/(B+C)+1}+{B/(C+A)+1}+{C/(A+B)+1}-3
\n" ); document.write( "S=(A+B+C){1/(B+C)+1/(C+A)+1/(A+B)}-3
\n" ); document.write( "WE HAVE B=A+1,C=A+2..SO
\n" ); document.write( "S=(A+A+1+A+2){1/(2A+3)+1/(2A+2)+1/(2A+1)}-3
\n" ); document.write( "S=3(A+1){1/(2A+3)+1/(2A+2)+1/(2A+1)}-3
\n" ); document.write( "SINCE A,B,C ARE POSITIVE INTEGERS MIIMUM VALUE OF A IS 1 AND MAXIMUM
\n" ); document.write( "IS TENDING TO INFINITY…..
\n" ); document.write( "IF WE PUT L=1/(2A+3)+1/(2A+2)+1/(2A+1),THEN WE FIND 1/(2A+1) IS THE
\n" ); document.write( "BIGGEST OF THE 3 FRACTIONS AND 1/(2A+3) IS THE SMALLEST OF THE 3
\n" ); document.write( "FRACTIONS.HENCE
\n" ); document.write( "L-MAX =3/(2A+1)…..AND….. L-MIN=3/(2A+3)
\n" ); document.write( "S-MAX=9(A+1)/(2A+1)-3….AND….S-MIN=9(A+1)/(2A+3)-3
\n" ); document.write( "IF WE TEST THESE AT A=1 AND A TENDING TO INFINITY WE FIND THAT
\n" ); document.write( "S-MAX=9*2/3-3=3…OR…TOWARDS INFINITY…S-MAX=9/2-3=3/2=1.5
\n" ); document.write( "S-MIN =9*2/5-3=3/5….OR..TOWARDS INFINITY…S-MIN=9/2-3=3/2
\n" ); document.write( "HENCE WE CAN CONCLUDE THAT S-MAX IS 3/2 IN THE LEAST OR 3 AT THE HIGHEST.
\n" ); document.write( "AND S- MIN IS NEVER LESSTHAN 3/5
\n" ); document.write( "OR 3/5<=S<=3/2
\n" ); document.write( "SO OBVIOUSLY 6/5<=S<=2 \n" ); document.write( "
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