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

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

(Part (a) was to show that \(\displaystyle G\leq M_1\) and that \(\displaystyle G=M_1\) only when \(\displaystyle x_1=x_2=\ldots =x_n\). 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?

 

[Euge]

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

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

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

(Part (a) was to show that \(\displaystyle G\leq M_1\) and that \(\displaystyle G=M_1\) only when \(\displaystyle x_1=x_2=\ldots =x_n\). 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

\(\displaystyle G(a_1,...,a_n) := (a_1\cdots a_n)^{\frac{1}{n}}\) and \(\displaystyle M_r(a_1,...,a_n) := \left(\frac{a_1^r + \cdots + a_n^r}{n}\right)^{\frac{1}{r}},\)

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

\(\displaystyle [G(x_1,\ldots,x_n)]^p = G(x_1^p,\ldots, x_n^p) < M_1(x_1^p,\ldots,x_n^p) = [M_p(x_1,\ldots,x_n)]^p,\)

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,

\(\displaystyle G(x_1,\ldots,x_n) < M_p(x_1,\ldots,x_n),\)

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

\(\displaystyle [G(x_1,\ldots,x_n)]^q = G(x_1^q,\ldots,x_n^q) < M_1(x_1^q,\ldots,x_n^q) = [M_q(x_1,...,x_n)]^q.\)

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

\(\displaystyle G(x_1,\ldots,x_n) > M_q(x_1,...,x_n),\)

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 \(\displaystyle M_p\). I'll do it now just for completeness. If \(\displaystyle p\) is a nonzero integer, then the \(\displaystyle p\)th-power mean of \(\displaystyle n\) positive real numbers \(\displaystyle x_1,x_2,\ldots, x_n\) is \(\displaystyle M_p=\left(\frac{x_1^p\ldots +x_n^p}{n}\right)^{1/p}\).