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Check my proof

  1. Mar 26, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that [itex]\frac{a+b}{2}\geq\sqrt{ab}[/itex] for [itex]0 < a \leq b[/itex]

    3. The attempt at a solution

    Since [itex]b \geq a[/itex] then [itex]b + a \geq 2a[/itex] and [itex]2b\geq a + b[/itex]
    That is [itex]\frac{a+b}{2}\geq a[/itex] and [itex]\frac{a+b}{2}\leq b[/itex].

    Since [itex]\frac{a+b}{2}\leq b[/itex] can i multiply [itex]\frac{a+b}{2}\geq a[/itex] with [itex]b[/itex]?

    If i can do that then [itex](\frac{a+b}{2})^{2}\geq ab[/itex] that is:
    [itex]\frac{a+b}{2}\geq \sqrt{ab}[/itex]


    On a side note if problem says prove something for all positive intigers a,b and c.
    That means that a>0 , b>0, c>0 but can i take from that, [itex]a\geq b\geq c > 0[/itex]?
     
  2. jcsd
  3. Mar 26, 2013 #2
    How did you go from [itex]b\frac{a+b}{2}\geq ab [/itex]to [itex](\frac{a+b}{2})^{2} \geq ab [/itex]?
    You could just square both the right and left side from the start, so that
    [itex]0.25a^{2}+0.5ab+0.25b^{2} \geq ab \leftrightarrow 0.25a^{2}+0.25b^{2} \geq 0.5 ab \leftrightarrow a^{2}+b^{2}-2ab \geq 0 \leftrightarrow (a-b)^{2} \geq [/itex]0 which obviously is true.
     
  4. Mar 26, 2013 #3
    As for your last question, that does not tell you very much. It tels you that a is larger than or equal to b which is larger than or equal to c which is larger than 0. A more concise way to express the stipulation is to write
    [itex]a,b,c \in N[/itex]
    or
    [itex]a,b,c \in Z^{+}[/itex]
     
  5. Mar 26, 2013 #4
    Well i thought that since b≥ (a+b)/2 and (a+b)/2 * b = ((a+b)/2)^2 (because it can be b = (a+b)/2)
     
  6. Mar 26, 2013 #5

    Mark44

    Staff: Mentor

    But you can't treat this as an equality, which is what you seem to be doing.
     
  7. Mar 26, 2013 #6

    Curious3141

    User Avatar
    Homework Helper

    You can't do this: the inequality signs face different ways.

    If ##a \geq b## and ##c \geq d##, then for positive a,b,c and d, you CAN assert that ##ac \geq bd##. If that's a ##\leq## sign, then the opposite applies (just switch all the signs). However, if the signs are different you cannot conclude anything from multiplying the two inequalities.

    The simplest way to approach this is to start by observing that ##(\sqrt{a} - \sqrt{b})^2 \geq 0## for all ##a,b \in \mathbb{R^+}## (positive real a and b).

    No you cannot infer that. You can only say that a, b and c are all greater than 0, but you cannot conclude anything about how they're ordered with respect to one another.

    However, if a, b and c are *arbitrary* positive integers, and the property you're required to prove is essentially symmetric in a, b, and c (i.e. switching the symbols around doesn't matter), then you can start by arguing "without loss of generality, let ##a \geq b \geq c > 0##" if it helps your proof. This is a common technique.
     
    Last edited: Mar 26, 2013
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