Is there a fundamental way to prove Generalized Mean is a increasing function Generalized mean:
$$M_k=\left(\sum_{i=1}^n\frac{{x_i}^k}{n}\right)^{\frac 1 k}$$ 
I try to prove $L=\ln\left(M_k\right)$ is increasing. $$\frac{d\left(L\right)}{dk}=\frac{\sum_{i=1}^n{x_i}^k\ln\left(x_i\right)}{\sum_{i=1}^n{x_i}^k}$$ and I stuck here. I visited wikipedia and it says this can be proved using Jesen's inequality. But I want a simple one(or complicated one, if one think using Jesen's inequality is more easier). 
Any help is going to be appreciated. Thanks
 A: With Jensen's inequality, it is indeed easy.


*

*In the first place, suppose $0 <k<\ell$, and consider the function $f$ defined by $\;f(x)=x^{\tfrac \ell k}$. As $\dfrac\ell k >1$, this function is convex, so by Jensen's inequality,
$$\biggl(\frac{x_1^k+\dots+x_n^k}n\biggr)^{\!\tfrac\ell k}\le\frac{\bigl(x_1^k\bigr)^{\tfrac\ell k}+\dots+\bigl(x_n^k\bigr)^{\tfrac\ell k}}n=\frac{x_1^\ell+\dots+x_n^\ell}n$$
Raising both sides to the power $\dfrac1\ell$, we obtain
$$ M_k(x_1,\dots,x_n)=\biggl(\frac{x_1^k+\dots+x_n^k}n\biggr)^{\!\tfrac1 k}\le\biggl(\frac{x_1^\ell+\dots+x_n^\ell}n\biggr)^{\!\tfrac1\ell} =M_\ell(x_1,\dots,x_n).$$

*Suppose now $k<0<\ell$, and set $k=-k'$. We'll prove
$$\Biggl(\frac{x_1^{-k'}+\dots+x_n^{-k'}}n\Biggr)^{\!\!-\tfrac1{k'}}\le\bigl(x_1\cdots x_n\bigr)^{\tfrac1n}\le\Biggl(\frac{x_1^{\ell}+\dots+x_n^{\ell}}n\Biggr)^{\!\!\tfrac1{\ell}}. $$
The left-hand inequality is equivalent to
$$\Biggl(\frac1n\biggl(\frac1{x_1^{k'}}+\dots+\frac1{x_n^{k'}}\biggr)\Biggr)^{\!\!-1}\le\Bigl(x_1^{k'}\cdots x_n^{k'}\Bigr)^{\tfrac1n}$$
which is the harmonic-geometric means inequality.


The right-hand inequality is equivalent to the arithmetic-geometric means inequality:
$$\bigl(x_1^{\ell}\cdots x_n^{\ell}\bigr)^{\tfrac1n}\le\frac{x_1^{\ell}+\dots+x_n^{\ell}}n. $$


*

*The last case $k<\ell<0$ is easy: setting $k=-k',\ \ell=-\ell'$, it steps back to the first case.

A: In addition to Bernard's solution, here is my proof using wikipedia notation for power mean: $\bar x(m) = ({1\over n}\sum_{i=1}^n x_i^m)^{1\over m}$

