Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    I am trying to show that if 0<a<1 and 0<b<1 and we know that a is less than b then:

     \frac{2ab+a}{b+a+2ab} &lt; \frac{1}{2}

    I have simplified the fractions to:

    \frac{2b+1}{1+2b+\frac{b}{a}}

    but I can't see how this is less than 0.5
    • PS Helper
    Offline

    14
    Your best bet is to multiply through by the denominators of the two fractions to get 2(2ab+a) &lt; b+a+4ab (we can do this since everything is positive). Why must this be true?
    • Thread Starter
    Offline

    0
    ReputationRep:
    That must be true because b<1?

    but is that not using the fact that it is less than a half without having shown it?
    Offline

    2
    ReputationRep:
    2ab + a < b

    Should be false. Take a = 1/2 and b arbitrary in (1/2,1).
    • Thread Starter
    Offline

    0
    ReputationRep:
    so do you mean that it the inequality in the op is not true?
    Offline

    2
    ReputationRep:
    (Original post by sonic7899)
    so do you mean that it the inequality in the op is not true?
    Unless 1/2 < 0 it's false.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by DeanK22)
    Unless 1/2 < 0 it's false.
    Erm I think I made a mistake in my workings. The fraction I should be showing is less than a half is actually:


    \frac{2ab+a}{4ab+a+b}
    • PS Helper
    Offline

    14
    (Original post by sonic7899)
    Erm I think I made a mistake in my workings. The fraction I should be showing is less than a half is actually:


    \frac{2ab+a}{4ab+a+b}
    That makes more sense. You need to use the fact that b&gt;a.
    Offline

    17
    ReputationRep:
    (Original post by sonic7899)
    Erm I think I made a mistake in my workings. The fraction I should be showing is less than a half is actually:


    \frac{2ab+a}{4ab+a+b}
    In which case, nuodai's suggestion is the way to go here. What is the original question?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by nuodai)
    That makes more sense. You need to use the fact that b&gt;a.
    Ah I think I have just got it. Is is just that a<b and so:

    \frac{2ab+a}{4ab+a+b} &lt; \frac{2ab+a}{4ab+a+a} = \frac{2ab+a}{2[2ab+a]} = \frac{1}{2}?
    • PS Helper
    Offline

    14
    (Original post by sonic7899)
    Ah I think I have just got it. Is is just that a<b and so:

    \frac{2ab+a}{4ab+a+b} &lt; \frac{2ab+a}{4ab+a+a} = \frac{2ab+a}{2[2ab+a]} = \frac{1}{2}?
    Seems fine to me
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Will you be richer or poorer than your parents?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.