Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: AM-GM Inequality Proof

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    From Spivak's Calculus, Chapter 2 Problem 22 Part A:

    Here, [tex]A_{n}[/tex] and [tex]G_{n}[/tex] stand for the arithmetic and geometric means respectively and [tex]a_{i}\geq 0[/tex] for [tex]i=1,\cdots,n[/tex].

    Suppose that [tex]a_{1} < A_{n}[/tex]. Then some [tex]a_{i}[/tex] satisfies [tex]a_{i} > A_{n}[/tex]; for convenience, say [tex]a_{2} > A_{n}[/tex]. Let [tex]a^{*}_{1} = A_{n}[/tex], and let [tex]a^{*}_{2} = a_{1} + a_{2} - a^{*}_{1}[/tex]. Show that

    [tex]a^{*}_{1} a^{*}_{2} \geq a_{1} a_{2}[/tex]

    Why does repeating this process enough times eventually prove that [tex]G_{n} \leq A_{n}[/tex]? (This is another place where it is a good exercise to provide a formal proof by induction, as well as an informal reason.) When does equality hold in the formula [tex]G_{n} \leq A_{n}[/tex]?

    2. Relevant equations
    3. The attempt at a solution

    The first part of the proof was easy:

    [tex]A^{2}_{n} - (a_{1} + a_{2}) A_{n} + a_{1} a_{2} = (A_{n} - a_{1})(A_{n} - a_{2}) < 0[/tex] since it was given that [tex]a_{1} < A_{n} < a_{2}[/tex]. Rearranging this leads to [tex]a_{1} a_{2} < a^{*}_1 a^{*}_2[/tex], which is what the problem asked for.

    It is also clear that equality holds when [tex]a_{1} = a_{2} = \cdots = a_{n} = A_{n}[/tex]. The part that is baffling me is filling in the rest of the logic of this proof to show that [tex]G_{n} \leq A_{n}[/tex]. I'm not exactly sure what process I'm suppose to "repeat" as stated in the problem, or how I'm supposed to turn it into a formal proof by induction. None of [tex]a^{*}_{i}[/tex] for [tex]i \geq 3[/tex] were defined anywhere in the problem. Any help would be appreciated.
  2. jcsd
  3. Sep 18, 2010 #2
    Hmm, well you basically already proved that the GM of the a_i's is less than the GM of the [itex]a^*_i[/itex]'s. But note that the AM of the a_i's is equal to the AM of the [itex]a^*_i[/itex]'s, so perhaps try to exploit this fact.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook