Discreet weighted mean inequality 
Let ${p_{1}},{p_{2}},\ldots,{p_{n}}$   and
   ${a_{1}},{a_{2}},\ldots,{a_{n}}$   be positive real numbers and let
   $r$   be a real number. Then for $r\ne0$  , we define
   ${M_{r}}(a,p)={\left({\frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+\cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+\cdots+{p_{n}}}}}\right)^{1/r}}$
   and for $r=0$  , we define
   ${M_{0}}(a,p)={\left({a_{1}^{{p_{1}}}a_{2}^{{p_{2}}}\cdots
 a_{n}^{{p_{n}}}}\right)}^{1/\sum\nolimits _{i=1}^{n}p_i}$  . Then prove that  $
 {M_{{k_{1}}}}(a,p)\geqslant{M_{{k_{2}}}}(a,p)  $ if $k_{1}\geqslant
 k_{2}$.

How to prove this generalized theorem? I have found this in a book without any proof. So can anyone show me?

Let $$w_{i} = \frac{p_{i}}{\sum_{i=1}^np_{i}}\implies\sum_{i=1}^nw_{i}=1 $$

we recall that by Holder inequality one has
  $$\sum_{i=1}^nw_{i}A_i B_i
\le \left(\sum_{i=1}^nw_{i}A_i^{\color{red}{q}}\right)^{\color{red}{1/q}}\left(\sum_{i=1}^nw_{i}B_i^{\color{red}{q'}}\right)^{\color{red}{1/q'}}~~~~~~{\color{red}{1/q}}+{\color{red}{1/q'}}=1 $$ with $q\ge1$

Therefore, for $k\ge r$ if we let $q=\frac{k}{r}\ge 1$ we have,  $$\color{blue}{M_r(a,p)}={\left({\frac{{{p_{1}}a_{1}^{r}+{p_{2}}a_{2}^{r}+\cdots+{p_{n}}a_{n}^{r}}}{{{p_{1}}+{p_{2}}+\cdots+{p_{n}}}}}\right)^{1/r}} =\left(\sum_{i=1}^nw_{i}a_{i}^{r}\right)^{1/r}
\\=\left(\sum_{i=1}^nw_{i}\left(a_{i}^{k}\right)^{\color{red}{r/k}}\right)^{1/r}=\left(\sum_{i=1}^nw_{i}\left(a_{i}^{k}\right)^{\color{red}{1/q}}\cdot\color{blue}{1}\right)^{1/r}
\\\overset{Holder}{\le}\left(\left(\sum_{i=1}^nw_{i}\left(a_{i}^{k}\right)^{\color{red}{q/q}}\right)^{\color{red}{1/q}}\left(\sum_{i=1}^nw_{i}1^{\color{red}{1/q'}}\right)^{\color{red}{q'/q}}\right)^{1/r}~~~~~~{\color{red}{1/q}}+{\color{red}{1/q'}}=1 
\\=\left(\sum_{i=1}^nw_{i}a_{i}^{k}\right)^{1/qr}=\left(\sum_{i=1}^nw_{i}a_{i}^{k}\right)^{1/k}=\color{blue}{M_k(a,p)}~~~~~~qr=k $$
 that is $$M_{k}(a,p)\geqslant M_{r}(a,p)~~~~~k\ge r$$
where 
