General Mean Inequality

[*Mahi*]

Define $G_k$ as:
\[ G_k = \sqrt[k]{\frac {\sum_{i=1} ^ n a_i^k} {n}} \].
Then , $G_m \geq G _ n$ if and only if $m \geq n$.

[sm.joty]

Is this inequality true from the opposite sides ?

[nafistiham]

here,it says if and only if or iff so it is true from both the sides

if $n\leq m$ then $G_{n}\leq G_{m}$

[*Mahi*]

And it is true for $k \in \mathbb Z$

[sourav das]

Power mean inequality:
I'm giving a general form of this General Mean Theorem:
For
$\sum_{i=1}^{n}t_i=1$
$ t_i,x_i> 0 $
if $r>s$ then,
\[\left ( \sum_{i=1}^{n}t_ix_i^r \right )^\frac{1}{r}\geq \left ( \sum_{i=1}^{n}t_ix_i^s \right )^\frac{1}{s}\]
and equality holds if and only if all $x_i$ are same.
For, inequality, the converse is true.
But Equality holds for two cases:
(i)r=s;
(ii)all $x_i$ is equal.
General Mean Theorem is just a special case of this theorem when all $t_i=\frac{1}{n}$
And a little correction for equality: Equality holds
(i)r=s;
(ii)all $x_i$ is equal.
It is actually Exercise:1.87 of new book.
Solution:

Define a function \[f(x)=x^\frac{r}{s}\]
As r>s
\[f{}''(x)=(\frac{r}{s})(\frac{r-s}{s})x^{(\frac{r-s}{s}-1)}>0\]
So,\[f(x)=x^\frac{r}{s}\] is convex.
and so,
\[\left ( \sum_{i=1}^{n}t_ix_i^r \right )^\frac{1}{r}= \left ( \sum_{i=1}^{n}t_i(x_i^s)^{\frac{r}{s}} \right )^\frac{1}{r}\geq \left ( (\sum_{i=1}^{n}t_ix_i^s)^\frac{r}{s} \right )^\frac{1}{r}= \left ( \sum_{i=1}^{n}t_ix_i^s \right )^\frac{1}{s}\]