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

An interesting inequality-question on the proof

  1. Sep 4, 2008 #1
    An interesting inequality--question on the proof...

    I am working on the following proof and have gotten about half way;

    If [tex]a_1, a_2, \ldots , a_n[/tex] are positive real numbers then

    [tex] \sqrt[n]{a_1 \cdots a_n} \leq \frac{a_1+a_2+\ldots +a_n}{n}[/tex]

    By induction, I started by showing it for n=2, which goes as follows.

    We must show that [tex]\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}[/tex]. So consider some s, a positive real number. If

    [tex]\sqrt{a_1 a_2} \leq s \Rightarrow \sqrt{a_1 a_2} \leq \frac{\sqrt{a_1a_2}+s}{2} \leq s[/tex]

    And since [tex]\sqrt{a_1a_2}<a_1+a_2-\sqrt{a_1a_2}[/tex], just let that equal s. Then it follows that,

    [tex]\sqrt{a_1a_2} \leq \frac{\sqrt{a_1a_2}+a_1+a_2-\sqrt{a_1a_2}}{2}[/tex]

    Where after simplification we have shown that [tex]\sqrt{a_1 a_2} \leq \frac{a_1+a_2}{2}[/tex] or in other words it is true for n=2.

    Now for the next part of the proof, I think we must show that if it is true for [tex]n=2^m[/tex] then it is also true for [tex]n=2^{m+1}[/tex]. And then we must also show it if [tex]2^m < n < 2^{m+1}[/tex], but this is where I am stumped. Sorry for the choppiness in my explanation. Thanks for any assistance, I appreciate it.
  2. jcsd
  3. Sep 4, 2008 #2
    Re: An interesting inequality--question on the proof...

    Eh, well the n = 2 case could be proved in an easier way just be multiplying both side by 2, squaring, rearranging, and factoring. It follows from the trivial inequality [tex]x^2 \geq 0[/tex]

    As for the induction, I'm not sure why you use n = 2^m since it works for any natural number n. Anyways AM-GM is not easy to prove by induction since you have to apply an algorithm that is also used in a proof without induction.
  4. Sep 5, 2008 #3


    User Avatar
    Science Advisor
    Homework Helper

    Re: An interesting inequality--question on the proof...

    Actually, one of the classical proofs of the AM-GM inequality is an inductive one. The proof is due to Cauchy, and it follows the main ideas in the OP, i.e. inducting on powers of 2.

    There was a thread on this sometime back, here is the link. Check it out and post back if you need further help.
  5. Sep 5, 2008 #4
    Re: An interesting inequality--question on the proof...

    Oh alright, I didn't realize this was such a famous proof (Cauchy). Thanks for alerting me to that reference.
  6. Sep 6, 2008 #5
    Re: An interesting inequality--question on the proof...

    It can also be derived from the inequality

    [tex]\exp(x)\geq 1+x[/tex] (1)

    This is trivial to prove. It follows from this that:

    [tex]\left\langle\exp(X)\right\rangle = \exp\left(\left\langle X\right \rangle\right)\left\langle\exp\left(X - \left\langle X\right \rangle\right)\right\rangle[/tex]

    Apply (1) to the last factor:

    [tex]\left\langle\exp\left(X - \left\langle X\right \rangle\right)\right\rangle\geq \left\langle 1 + X - \left\langle X\right \rangle\right\rangle = 1[/tex]

    So, we have:

    [tex]\left\langle\exp(X)\right\rangle \geq \exp\left(\left\langle X\right \rangle\right)[/tex] (2)

    This is a special case of the convex inequality and the AM-GM inequality is a straightforward consequence of that inequality. But the convex inequality requires a bit more work to prove. The AM-GM inequality follows from (2) by taking X to be the logarithms of the a_i. The average on the l.h.s. is then the arithmetic mean, while on the r.h.s. you get the geometric mean.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook