# Direct proof of inequality between arithmetic and harmonic mean

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 :)

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

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 $\,H(a_1,\dots,a_n)$.
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 $n$.