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Inequality proof

  1. Mar 22, 2010 #1
    Sorry for not being very descriptive in the title, but here's my problem:

    Prove that for all [tex]a,b,c,d \ge 0[/tex], the following inequality is valid:

    [tex]\frac{a+b+c+d}{4} \ge \sqrt[4]{abcd}[/tex]

    I've tried to make it this way:

    [tex](x-y)^2 \ge 0[/tex]
    [tex]x^2 + y^2 \ge 2xy[/tex]


    [tex][(a+b)-(c+d)]^4 \ge 0[/tex]
    [tex][ (a+b)^2 - 2(a+b)(c+d) + (c+d)^2 ] ^2 \ge 0[/tex]
    [tex](a+b)^2 + (c+d)^2 \ge 2(a+b)(c+d)[/tex]
    [tex](a+b)^2 + 2(a+b)(c+d) + (c+d)^2 \ge 4(a+b)(c+d)[/tex]
    [tex]a+b+c+d \ge 2 \sqrt{(a+b)(c+d)}[/tex]
    [tex]\frac{a+b+c+d}{2} \ge \sqrt{(a+b)(c+d)}[/tex]

    This is the farthest I got, can't figure out how to continue... any tips?



    Prove, by induction, that the following inequality is true:

    [tex]\prod_{n=1} \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}[/tex]

    I made this:

    [tex]\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{1}{\sqrt{2n+1}} \cdot \frac{2n}{2n+1}[/tex]

    [tex]\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{2n}{\sqrt{(2n+1)(2n+1)^2}}[/tex]

    [tex]\left ( \prod_{n=1} \frac{2n-1}{2n} \right ) \cdot \frac{2n}{2n+1} < \frac{2n}{\sqrt{(2n+1)^3}}[/tex]

    Though I am really new to induction, so I'm not sure if my proof is finished.
    Last edited: Mar 22, 2010
  2. jcsd
  3. Mar 22, 2010 #2


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    The notation for your induction proof is terrible. You should be doing something like multiplying from n=1 to N, and then adding on [tex]\frac{2N}{2N+1}[/tex] at the end. As written, the left hand side of each of those equations doesn't really make any sense (it's like saying [tex] (\sum_{i=0}i^2) + (i+1)^2[/tex]... i is a dummy variable, you can't use it outside the summation like this. What you really want to write in this case is [tex] (\sum_{i=0}^n i^2) + (n+1)^2[/tex]
  4. Mar 22, 2010 #3
    Ok, I tried to abbreviate the notation, but apparently I failed miserably.


    [tex]\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdots \frac{2n-1}{2n} < \frac{1}{\sqrt{2n+1}}[/tex]
  5. Mar 22, 2010 #4
    Do you know the Arithmetic-Geometric Inequality?

    http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means" [Broken]
    Last edited by a moderator: May 4, 2017
  6. Apr 5, 2010 #5
    From [itex]x^2+y^2 \ge 2xy[/itex], consider adding [itex]2xy[/itex] to both sides of the inequality. You should quickly see that

    [tex]\frac{x+y}{2} > \sqrt{xy}[/tex]

    Now try apply the above to [itex]x=\sqrt{ab}[/itex] and [itex]y=\sqrt{cd}[/itex].
  7. Apr 6, 2010 #6


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    Yeah this is only AM-GM for four numbers, but you can also use Holder's inequality or Jensen's inequality. In fact, AM-GM inequality is just a corollary of Jensen's inequality (hint- for proof consider the function f(x)=ln x). As for AM-GM inequality, it is:
    If [tex]a_i , i=1,2,...,n[/tex] are positive real numbers then:
    [tex] \frac{a_1 +a_2 +...+a_n}{n} \geq \sqrt[n]{a_1 a_2 ...a_n} [/tex]
    This can be generalized into something that we call General Mean Inequality.

    Anyways, To prove this you can use induction on [tex] n [/tex]. However, to prove the inequality that you posted, you may directly see that if the inequality holds for [tex]n=2 [/tex] (for which you have the proof) then obviously holds for every [tex]n[/tex] which is a power of [tex] 2 [/tex].

    As for you second inequality, take the [tex] ln [/tex] of both sides, and then try to use Jensen's inequality (hint- what can you say about the second derivative of f(x)=ln x ? )
  8. Apr 6, 2010 #7
    Holder's and Jensen's inequalities will also do it, but it is much easier to understand using only elementary arguments. Just apply

    \frac{x+y}{2} > \sqrt{xy}

    sufficiently many times (namely with [itex]x=\sqrt{ab}[/itex] and [itex]y=\sqrt{cd}[/itex]). What is [itex]\sqrt{xy}[/itex] in terms of a, b, c, and d? What is [itex]\frac{x+y}{2}[/itex] in terms of a, b, c, and d? Finally, consider applying the inequality above for [itex]\sqrt{ab}[/itex] (and similarly for [itex]\sqrt{cd}[/itex]). This will not only give you the desired inequality, it will also give you insight into the inductive step of the general inequality.
    Last edited: Apr 6, 2010
  9. Apr 6, 2010 #8


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    @rs1n: Can you post/direct your proof of General Mean Inequality using induction.

    Personally, I am aware of only one proof for it, it is simply using the fact that:
    [tex]f(n)=\left(\frac{a_1 ^n +a_2 ^n +...+a_n ^n}{n}\right)^{1/n} [/tex] is increasing function, for [tex] a_i [/tex] being positive.
    Last edited: Apr 6, 2010
  10. Apr 6, 2010 #9
    Sorry, my mind must have fixated on the phrase "general mean inequality" when I actually meant general inequality -- as in for any n, not just n=4 -- in my response. As for the GMI, the only proof I know is likely the same as that one you are familiar with.
  11. Apr 6, 2010 #10
    The original (I believe) proof was by induction, but starting with powers of 2 and then filling in the rest. That proof is right on the wikipedia page for AM-GM.

    edit: Oops, looks like I can't read. I don't know if this technique can be used for the more general inequality you guys are talking about.
    Last edited: Apr 6, 2010
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