A.M.>G.M. of four numbers

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Solution 1

Hint: If you can use the AM-GM on 2 numbers, then you should use that.

First prove with this inequality on $(a,b)$ $$\frac{a+b}{2} \geq \sqrt{ab}$$

Then prove with this inequality on $(c,d)$ $$\frac{c+d}{2} \geq \sqrt{cd}$$

Then use the AM-GM on 2 numbers again on the terms above.

Solution 2

The inequality $\text{AM}>\text{GM}$ is equivalent to $\log\text{AM}>\log\text{GM}$, because the logarithm is a monotone increasing function.

Expanding $\log\text{AM}$ and $\log\text{GM}$,

\begin{align} \log\text{AM} & = \log \frac{a+b+c+d}{4} \\ \log\text{GM} & = \log(abcd)^{1/4} = \frac{\log a+\log b+\log c+\log d}{4} \end{align}

The result then follows because the logarithm is a concave function (see Jensen's inequality)

Solution 3

we have by AM-GM: $$\frac{\frac{a+b}{2}+\frac{c+d}{2}}{2}\geq \sqrt{\frac{a+b}{2}\frac{c+d}{2}}\geq \sqrt{\sqrt{ab}\sqrt{cd}}$$

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Wisha
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Wisha

Updated on February 07, 2020

Comments

  • Wisha
    Wisha over 3 years

    Prove that arithmetic mean of $4$ numbers is greater than geometric mean of the same $4$ numbers, i.e. prove that $$\dfrac{a+b+c+d}{4} > (abcd)^{\frac1{4}}$$

    • 5xum
      5xum almost 8 years
      Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it?
    • Ruben
      Ruben almost 8 years
      For correctness, I think you should add that a, b, c, d should all be different and non-negative.
  • Thomas Ahle
    Thomas Ahle almost 8 years
    I don't think your expansion is correct. You need to divide everything by $4^4$.
  • NeilRoy
    NeilRoy almost 8 years
    Sorry ... my bad!
  • wythagoras
    wythagoras almost 8 years
    This is true by Jensen's inequality.
  • Luis Mendo
    Luis Mendo almost 8 years
    @wythagoras Thanks. I've included that
  • Ruben
    Ruben almost 8 years
    $ 4^4 $ > 24, making the proof incorrect. Would have been a nice proof though!
  • wythagoras
    wythagoras almost 8 years
    Well you could make all kinds of horrible estimates (e.g $a^4+b^4+c^4+d^4>4abcd$ by the rearrangement inequality) but I wouldn't recommend that.
  • Wisha
    Wisha almost 8 years
    hi your method is interesing.
  • Wisha
    Wisha almost 8 years
    can u pls tell me what is concave function?
  • Luis Mendo
    Luis Mendo almost 8 years
    It's a function that lies above any of its chords. It's the opposite of a convex function. Did you follow the link in my question?
  • NeilRoy
    NeilRoy almost 8 years
    @MonKeePoo yeh thats right... but if we shift $4^4=256$ to rhs...the lhs will still be huge!
  • Jolien
    Jolien almost 8 years
    Why is this clearly bigger than $abcd$? It could be the case that for example $4c^3d<0$.