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Abstract Inequalities

  1. Oct 19, 2012 #1
    1. The problem statement, all variables and given/known data

    The final answer I have of [itex](a+b)(a-b)[/itex] does not appear to fit the text book's required "results of inequalities which hold true for all real no.s", i.e. either: 1. [itex](a)^2[/itex] or [itex](a-b)^2[/itex] or 2. [itex]-(a+b)^2[/itex]. Can anyone confirm if I have solved this correctly, in line with the conditions the book has described above?

    Many thanks.

    Q. If a > 0 & b > 0, show that: [itex]\frac{1}{a}+\frac{1}{b}\geq\frac{2}{a+b}[/itex]


    2. Relevant equations


    3. The attempt at a solution

    Attempt: if [itex]\frac{a+b}{ab}\geq\frac{2}{a+b}[/itex]
    if [itex]a+b\geq\frac{2ab}{a+b}[/itex]
    if [itex](a+b)(a+b)\geq2ab[/itex]
    if [itex]a^2+2ab+b^2-2ab\geq0[/itex]
    if [itex](a+b)(a-b)\geq0[/itex]
     
  2. jcsd
  3. Oct 19, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    It should be obvious that (a+ b)(a- b) is NOT always positive: take a= 1, b= 2 so that (a+ b)(a- b)= (3)(-1)= -3.
     
  4. Oct 19, 2012 #3
    In which case I should conclude with:
    if [itex]a^2+b^2\geq0[/itex]?
     
  5. Oct 19, 2012 #4

    Mark44

    Staff: Mentor

    Why do you have "if" starting each line?

    You have done several things that aren't always valid.
    1. You got to the 2nd step by multiplying both sides of the original inequality by ab. If a and b have opposite signs, the inequality direction in step 2 needs to switch.
    2. To get to step 3, you multiplied by a + b. If a + b < 0, the inequality direction needs to switch.

    You haven't taken this into consideration, so what you ended with doesn't necessarily follow what you started with.
     
  6. Oct 22, 2012 #5
    To account for the 'if's', I'll post an image of an example from the book where I got this question from (apologies for the slightly fuzzy image).
    As for points 1. & 2., I guess I will need to keep the clutter away from the RHS of the equation, so the inequality remains true, and I avoid having to change signs.
     

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