Discreet weighted mean inequality

Question

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?

Check here: http://math.stackexchange.com/questions/922401/my-favorite-proof-of-the-generalized-am-gm-inequality-where-it-came-from A proof is presened by the OP. — MathematicianByMistake, May 09 '16 at 07:18
Jensen's inequality comes to mind. Have a look : https://en.wikipedia.org/wiki/Jensen%27s_inequality — Aritra Das, May 09 '16 at 07:18
This is more than the generalized AM GM. It includes all weighted means. — Perth, May 09 '16 at 07:41

score 1 · Answer 1 · answered Dec 19 '17 at 17:53

In fact this inequality is a generalized Radon inequality present in this paper : A note on the proofs of generalized Radon inequality. See the Theorem 2.2 for a proof . This paper is very interesting because we can see the link between differents inequalities like :

(i) Bernoulli inequality,

(ii) the weighted AM-GM inequality,

(iii) Holder inequality,

(iv) the weighted power mean inequality,

(v) Minkovski inequality,

(vi) Radon inequality.

See the theorem 2.3 for that .

why do you need all that? Holder inequality or Jessen inequality is sufficiently enough — Guy Fsone, Dec 19 '17 at 18:22

Guy Fsone · Answer 2 · 2017-12-19T18:32:46.337

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

Discreet weighted mean inequality

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