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Conditional Proof question

  1. Nov 7, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##x\in \mathbb{R} ##
    Prove the conditional statement that,
    if ## x>-1## then ## x^2 + \frac {1}{x^2+1} \geq 1##

    2. The attempt at a solution

    Suppose ## x>-1## is true.
    Then ## x^2>1##
    Then ## \frac{1}{2}>\frac {1}{x^2+1}##
    Then ##x^2+ \frac{1}{2}>x^2+\frac {1}{x^2+1}##

    After that I have no clue how to get to the part where ## x^2 + \frac {1}{x^2+1} \geq 1## happens. Pls help...
     
  2. jcsd
  3. Nov 7, 2016 #2

    fresh_42

    Staff: Mentor

    Aren't you already done if ##x^2 \geq 1\;##? Isn't ##\frac{1}{1+x^2}## simply something positive?
    But ##x > -1## doesn't imply ##x^2 > 1##, e.g. ##x = -\frac{1}{2}##.
     
  4. Nov 7, 2016 #3
    Yup you're correct, but then how do I do it. I have no idea how to get something like ##x^2\geq1## from ## x>-1##.
     
  5. Nov 7, 2016 #4

    fresh_42

    Staff: Mentor

    You can distinguish between the two cases.
    1. ##|x| \geq 1##
    2. ##|x| < 1##
    In the first case you have (together with ##x >-1##) that ##x \geq 1## and ##x^2 \geq 1##.
    In the second case, have a look on ##\frac{1}{x^2+1}## first.
     
  6. Nov 7, 2016 #5
    Thank you...
     
  7. Nov 7, 2016 #6

    Math_QED

    User Avatar
    Homework Helper

    Deleted
     
  8. Nov 7, 2016 #7

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Is the restriction ##x >-1## important for getting the inequality?
     
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