MHB Elementary proof of generalized power mean inequality

[Ragnarok7]

This is problem 20b from chapter I 4.10 of Apostol's Calculus I.

The geometric mean $$G$$ of $$n$$ positive real numbers $$x_1,\ldots, x_n$$ is defined by the formula $$G=(x_1x_2\ldots x_n)^{1/n}$$.

Let $$p$$ and $$q$$ be integers, $$q<0<p$$. From part (a) deduce that $$M_q<G<M_p$$ when $$x_1,x_2,\ldots, x_n$$ are not all equal.

(Part (a) was to show that $$ and that $$ only when $$. I.e., the AM-GM inequality.)

Generalized mean - Wikipedia, the free encyclopedia

Wikipedia mentions it is proved using Jensen's inequality, and all the proofs I have found have used this. But since this comes from the opening of Apostol's book, where calculus has not been developed, I'm wondering if there is a non-calculus way of proving this. Does anyone have any ideas?

 

[Euge]

This is problem 20b from chapter I 4.10 of Apostol's Calculus I.

The geometric mean $$G$$ of $$n$$ positive real numbers $$x_1,\ldots, x_n$$ is defined by the formula $$G=(x_1x_2\ldots x_n)^{1/n}$$.

Let $$p$$ and $$q$$ be integers, $$q<0<p$$. From part (a) deduce that $$M_q<G<M_p$$ when $$x_1,x_2,\ldots, x_n$$ are not all equal.

(Part (a) was to show that $$G\leq M_1$$ and that $$G=M_1$$ only when $$x_1=x_2=\ldots =x_n$$. I.e., the AM-GM inequality.)

Generalized mean - Wikipedia, the free encyclopedia

Wikipedia mentions it is proved using Jensen's inequality, and all the proofs I have found have used this. But since this comes from the opening of Apostol's book, where calculus has not been developed, I'm wondering if there is a non-calculus way of proving this. Does anyone have any ideas?

Hi Ragnarok,

To prove the result, we must show that $G < M_p$ and $G > M_q$ when $x_1\ldots,x_n$ are not all equal. For definiteness, I'll let

$$ and $$

for all n-tuples $(a_1,\ldots a_n)$ of positive real numbers. Then

$$

where the first inequality follows from part (a) and the fact that $x_1^p,\ldots,x_n^p$ are not all equal. Hence, taking $p^{th}$-roots,

$$

or $G < M_p$. To prove $G > M_q$, we use a similar argument:

$$

The strict inequality is due to part (a) and the fact that $x_1^q,\ldots, x_n^q$ are not all equal. Since $-q > 0$, we take $(-q)^{th}$ roots and then reciprocate to obtain

$$

i.e., $G > M_q$.

 

[Ragnarok7]

Thank you so much Euge! That is a great proof and I didn't think it could be so simple!

I just realized that I forgot to define $$. I'll do it now just for completeness. If $$ is a nonzero integer, then the $$th-power mean of $$ positive real numbers $$ is $$.