0
$\begingroup$

I need to prove inequality from the title. I know that it follows from $H_n \leq G_n \leq A_n \leq Q_n$, where $H_n, G_n, A_n, Q_n$ are harmonic, geometric, arithmetic and quadratic means of $n$ real numbers, but for some purpose, I need to prove directly that $A_n \geq H_n$.

I have searched and couldn't find anything similar.

EDIT: I forgot to say that I have found this, but it uses Cauchy's Inequality. I would like to find some proof without it, as it exceeds level of the paper I'm writing :)

$\endgroup$
  • $\begingroup$ proofwiki.org/wiki/… $\endgroup$ – Zev Chonoles Jun 18 '15 at 11:34
  • $\begingroup$ Thank you, I forgot to say that I found that, but that's not what I need as it uses Cauchy's inequality. $\endgroup$ – Steph Jun 18 '15 at 11:43
2
$\begingroup$

Use Cauchy-Schwarz inequality: \begin{align*}&\biggl(\bigl(\sqrt a_1,\dots,\sqrt a_n\bigr)\cdot\biggl(\frac1{\sqrt a_1},\dots,\frac1{\sqrt a_n}\biggr)\biggr)^2=n^2\le\bigl(a_1+\dots+a_n\bigr)\Bigl(\frac1{a_1}+\dotsm\frac1{a_n}\Bigr)\\ \iff & \frac n{\cfrac1{a_1}+\dotsm\cfrac1{a_n}}\le \frac{a_1+\dots+a_n}n. \end{align*} The l.h.s. of the last inequality is exactly $\,H(a_1,\dots,a_n)$.

$\endgroup$
0
$\begingroup$

So we have to prove that $a_i>0$ gives: $$\frac{a_1+\ldots+a_n}{n}\leq \frac{n}{\frac{1}{a_1}+\ldots+\frac{1}{a_n}}\tag{1}$$ but by Titu's lemma: $$\left(\frac{b_1^2}{a_1}+\ldots+\frac{b_n^2}{a_n}\right)\left(a_1+\ldots+a_n\right)\geq (b_1+\ldots+b_n)^2 \tag{2}$$ so $(1)$ trivially follows by taking $b_i=1$. $(2)$ can easily be proved by induction on $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.