Im doing some research as a final project for my mathematics degree in the university and I can not find a proof of the following inequality, involving Theory of Means, could someone do it?
If A is the aritmetic mean, G is the geometric mean and H is the harmonic mean of a set of values $x=(x_1,x_2,\ldots,x_n)$, prove that
$H^{n-1}(x)A(x) \leq G^n(x) \leq A^{n-1}(x)H(x)$
So far I know that this inequality is called "The Sierpinski inequality". But I can not find a proof.
This is the so-called Sierpinski inequality.
Here is a copy of the (recursive) proof from the "Handbook of means and their inequalities", P.S. Bullen, North Holland Ed. 1988, pages 150-151.
Notations: Gothic letters $\frak{A_n,G_n,H_n}$ for $A,G,H$ and $\underline{a}=(a_1, a_2, \cdots a_n)$.
(HA) means Harmonic-arithmetic means inequality.
Edit : The following elements should be added
For the case $n=2$: $$G(a)^2=H(a)A(a) \iff ab=\frac{2}{\tfrac1a+\tfrac1b}\frac{a+b}{2}$$
One needs only to prove the second inequality in (57). Indeed, the first inequality is obtained by replacing sequence $a_i$ by sequence $\frac{1}{a_i}$, due to the fact that :
$$H_n(a)=\dfrac{n}{\tfrac{1}{a_1}+\tfrac{1}{a_1}+\cdots + \tfrac{1}{a_n}}=\dfrac{1}{A_n(\tfrac{1}{a})} \ \text{and} \ G_n(a)=\dfrac{1}{G_n(\tfrac{1}{a})}$$
Caution: There is an error in the formula upwards for $x'$ which should be with a denominator $n-2$ instead of $n-1$. I have noticed this error while trying to understand the issue by considering the case $n=4$ with fixed values $a,b,c$ and a variable. Here is a Geogebra representation of function $g$ (blue colored curve). Function denoted $f$ (red colored curve) in this representation is the derivative of $g$. It appears that setting $g'(x)=0$ gives a first degree equation $Ax+B=0$ with $B<0$ and $A>0$ explaining that $x'>0$.