inequality proof by calculus

[abcalphaomega]

can someone pls give me a link or some proof of why the function [(a^n+b^n+c^n)/3]^1/n is monotonically increasing for n>1. i can see that am-gm, am-qm etc follow from this fact. also pf for mean of k nos will be preferable than for 3 stated above.

 

also can i replace sigma a_iⁿ / k by (definite integral from a to b xⁿ dx) / b-a in the above inequality??

[timo]

- Proper capitalization kicks ass!

- What variable is the term a function of?

- What is "am-gm" and "am-qm" supposed to mean?

- "Also pf for mean of k nos will be preferable than for 3 stated above" ...

 

In short: Try being at least remotely comprehensible - especially with a question about math. Precise statements are the essence of this science.

[abcalphaomega]

sorry was in a bit of hurry last time.

am-gm means AM- GM inequality similarly AM - quadratic mean (am-qm)inequality.

the function is a variable of n.

im simply looking for a proof of

f(n)= [ a^n +b^n + c^n / 3 ]^(1/n) is monotonically increasing for n>1. eg. f(2)>f(1) implies AM-QM.

 

 

- Proper capitalization kicks ass!

- What variable is the term a function of?

- What is "am-gm" and "am-qm" supposed to mean?

- "Also pf for mean of k nos will be preferable than for 3 stated above" ...

 

In short: Try being at least remotely comprehensible - especially with a question about math. Precise statements are the essence of this science.

 

also i might add a,b,c are positive real nos.

 

incidentally i stumbled upon the need for such an inequality in trying to prove that for a given volume sphere has minimum surface area. something often told in fluid mechanics classes. i will be done with the proof if i find an analogous proof as that of the above inequality only involving averaged integration of a function rather than mean of discrete quantities.

[Xittenn]

Wouldn't the proof for AM-QM Inequality still hold for this generalized form? Or is this what you are trying to give proof of?

 

latex is pretty :/

 

[math] f(n) = \sqrt[n]{ \frac{x^n_1+x^n_2+x^n_3.....x^n_i}{i} } [/math]

Edited February 13, 2011 by Xittenn

[abcalphaomega]

yes, this exactly...

 

Wouldn't the proof for AM-QM Inequality still hold for this generalized form? Or is this what you are trying to give proof of?

 

latex is pretty :/

 

[math] f(n) = \sqrt[n]{ \frac{x^n_1+x^n_2+x^n_3.....x^n_i}{i} } [/math]

[timo]

Did you try looking at df/dn ?

[Xittenn]

I might be missing something but this seems pretty clear:

 

AM-QM states that:

 

[math] \frac{x_1+x_2+x_3.....x_i}{i} \leq \sqrt[2]{ \frac{x^2_1+x^2_2+x^2_3.....x^2_i}{i} } [/math]

 

which is like

 

[math] \sqrt[1]{ \frac{x^1_1+x^1_2+x^1_3.....x^1_i}{i} } \leq \sqrt[2]{ \frac{x^2_1+x^2_2+x^2_3.....x^2_i}{i} } [/math]

 

and I would assume by recursion that:

 

[math] \sqrt[n-1]{ \frac{x^{n-1}_1+x^{n-1}_2+x^{n-1}_3.....x^{n-1}_i}{i} } \leq \sqrt[n]{ \frac{x^n_1+x^n_2+x^n_3.....x^n_i}{i} } [/math]

 

did I miss something .... the proof for AM-QM is readily found online I'm sure ... forgive me if I stuck my foot in my mouth

Edited February 14, 2011 by Xittenn

[abcalphaomega]

yes i tried looking at df/dn. i dont see why the expression must be greater than 0 for n>1. its not very neat as it is.

[Xittenn]

http://mathstat.helsinki.fi/EMIS/journals/JIPAM/v3n3/014_02.pdf

 

I like this one because Guelph is a great place to party .... not because it is an uber testament to mathematical genius. Regardless it is a proof of Power Means Inequality, I must thank you for making me look. Other sources include:

 

http://www.emis.de/journals/JIPAM/images/029_04_JIPAM/029_04.pdf

 

http://en.wikipedia.org/wiki/Generalized_mean#Proof_of_power_means_inequality

 

Matrix Mathematics by Dennis Bernstein

 

Ok avoid the first one ... but still

[abcalphaomega]

thanks a million this was exactly what i was looking for.