Let . Show that for real , . 3 Responses to Range finder Pratik Poddar had this to say about that: October 13, 2014 at 6:03 am x^2+x+1/4 = (x+1/2)^2>=0 x^2+x+2>=1.75 { Adding 1.75 on both sides } 0<=1/(x^2+x+2)= -1/(x^2+x+2)>= -4/7 { Multiplying by -1 } 1>= (x^2+x+1)/(x^2+x+2)>=3/7 { Adding 1 } Hence proved ðŸ˜› Pratik Poddar had this to say about that: October 13, 2014 at 6:05 am Some typing error in my last comment x^2+x+1/4 = (x+1/2)^2>=0 x^2+x+2>=1.75 { Adding 1.75 on both sides } 0<=1/(x^2+x+2)=-1/(x^2+x+2)>= -4/7 { Multiplying by -1 } 1>= (x^2+x+1)/(x^2+x+2)>=3/7 { Adding 1 } Hence proved ðŸ˜› Pratik Poddar had this to say about that: October 13, 2014 at 6:09 am Your template is not accepting my answer. gte is greater than or equal to, and lte is less than or equal to, and eq is equals. x^2+x+1/4 eq (x+1/2)^2 gte 0 x^2+x+2 gte 1.75 { Adding 1.75 on both sides } 0 lte 1/(x^2+x+2) lte 4/7 { Taking reciprocal } 0 gte -1/(x^2+x+2) gte -4/7 { Multiplying by -1 } 1 gte (x^2+x+1)/(x^2+x+2) gte 3/7 { Adding 1 } Hence proved ðŸ˜› Subscribe to comments

x^2+x+1/4 = (x+1/2)^2>=0

x^2+x+2>=1.75 { Adding 1.75 on both sides }

0<=1/(x^2+x+2)= -1/(x^2+x+2)>= -4/7 { Multiplying by -1 }

1>= (x^2+x+1)/(x^2+x+2)>=3/7 { Adding 1 }

Hence proved ðŸ˜›

Some typing error in my last comment

x^2+x+1/4 = (x+1/2)^2>=0

x^2+x+2>=1.75 { Adding 1.75 on both sides }

0<=1/(x^2+x+2)=-1/(x^2+x+2)>= -4/7 { Multiplying by -1 }

1>= (x^2+x+1)/(x^2+x+2)>=3/7 { Adding 1 }

Hence proved ðŸ˜›

Your template is not accepting my answer. gte is greater than or equal to, and lte is less than or equal to, and eq is equals.

x^2+x+1/4 eq (x+1/2)^2 gte 0

x^2+x+2 gte 1.75 { Adding 1.75 on both sides }

0 lte 1/(x^2+x+2) lte 4/7 { Taking reciprocal }

0 gte -1/(x^2+x+2) gte -4/7 { Multiplying by -1 }

1 gte (x^2+x+1)/(x^2+x+2) gte 3/7 { Adding 1 }

Hence proved ðŸ˜›