QM-AM-GM-HM proof help 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...
 A: 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
$$A_n=\frac1{n}\sum_{i=1}^{n}x_i.$$
A: 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 $y_i = 1/x_i$.
