QM-AM-GM-HM proof help

Question

Out of interest, I am trying to proof QM-AM-GM-HM inequality. If you don't know it, it's something like this...
Let there be $n$ numbers $x_1, x_2, x_3...x_n$, where $x_1, x_2, ...,x_n>0$.
Proof that $$\sqrt{\frac{x_1^2+x_2^2...+x_n^2}{n}}\geqslant{\frac{x_1+x_2...+x_n}{n}}\geqslant{\sqrt[n]{x_1x_2...x_n}}\geqslant{\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}...+\frac{1}{x_n}}}$$ I thought of using induction (for n). The base case was something that took me about 20 mins to solve. I used n=2 (n=1 was trivial) but I am stuck. Can anyone give me a hint to continue me? To be exact, I need help in apply the induction hypothesis to the induction step. The numbers/fractions are starting to get ... uh ... ugly... Update 1: I don't want to see the answer. Just a hint...

Some advice; typically the nice thing about the base case is that it should be trivial to prove, so there's nothing wrong in picking $n=1$ as your base case. Times when doing a later base case like $n=2$ are when this is used for the induction hypothesis. — Hayden, Jun 20 '14 at 14:12
What does QM stand for? (Obviously it is not Quantum Mechanics; ??? mean?) — KCd, Jun 20 '14 at 14:13
AM-GM inequality is a nice example of result where Cauchy induction can be used. We also have a list of proofs of AM-GM inequality on this site. — Martin Sleziak, Jun 20 '14 at 14:20

RRL · Answer 1 · 2014-06-20T14:23:38.810

1

Hint for AM-GM:

Note that

$$(x_1x_2)(x_3x_4)\leq\left[\frac{x_1+x_2}{2}\right]^2\left[\frac{x_3+x_4}{2}\right]^2 \leq \left[\frac1{4}\sum_{i=1}^{4}x_i\right]^4.$$

Use this to prove by induction that

$$\left[\prod_{i=1}^{2^n}x_i\right]^{1/2^n} \leq \frac1{2^n}\sum_{i=1}^{2^n}x_i.$$

If $n$ is not a power of $2$, then choose $m$ such that $n+q=2^m$ and apply the previous result to$x_1,x_2,\ldots,x_n,A_n,\ldots,A_n$ where $A_n$ is repeated $q$ times and is the arithmetic average

$$A_n=\frac1{n}\sum_{i=1}^{n}x_i.$$

edited Jun 20 '14 at 14:23

answered Jun 20 '14 at 14:15

RRL

90,707

ok. seems convincing. What about the others? Roughly, I get the idea, but are there other methods? – Jun 20 '14 at 14:27
There are -- as others are showing -- but that is the straightforward induction approach. The far right inequality follows directly from AM-GM: switch $x_i$ with $1/x_i$ – RRL Jun 20 '14 at 14:41

S.B. · Accepted Answer · 2014-06-20T14:54:54.430

1

If you're not restricted to proof by induction, you can try to show that $$ M(p; x_1,x_2,\dotsc,x_n) := \left(\frac{1}{n}\sum _{i=1} ^n x_i^p\right)^{1/p},$$ is an increasing function of $p\in\mathbb{R}$. You only need Jensen's inequality to prove this.

update: For proof without calculus, you only need to prove the AM-GM inequality (e.g., through the Cauchy induction as others suggested). QM-AM is a simple case of the Cauchy-Schwarz inequality (which has an elementary proof). Furthermore, GM-HM is the same as AM-GM for the numbers $y_i = 1/x_i$.

edited Jun 20 '14 at 14:54

answered Jun 20 '14 at 14:34

S.B.

3,285

I don't really know the notation... – Jun 20 '14 at 14:36
@jonnytan999 What notation? I merely defined a function $M(p)$. – S.B. Jun 20 '14 at 14:37
QM = $M(2)$, AM = $M(1)$, GM = $\lim_{p\to 0} M(p)$, HM = $M(-1)$. – S.B. Jun 20 '14 at 14:41
you mean that M(p)= that, correct? – Jun 20 '14 at 14:41
@jonnytan999 Yes, as I mentioned particular values of $M(p)$ give you the QM, AM, GM, and HM of the $x_i$s. – S.B. Jun 20 '14 at 14:46
@jonnytan999 The way I know to prove this claim, requires taking one derivative, though. Do you need an elementary solution (i.e., without any calculus involved)? – S.B. Jun 20 '14 at 14:49
not really... depends on how complex that derivative will be... – Jun 20 '14 at 14:55

QM-AM-GM-HM proof help

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