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I'm doing a three part question to prove that $(a_1+a_2+a_3+...+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}\right)≥n^2$ for $a_1, a_2, ..., a_n\in\Bbb{R^+}$

(i) Assume Bernoulli's inequality, that $(1+x)^r\geq1+rx$ for all $x\geq-1$ and $r\in{\Bbb{Z}} \cap r\geq0$, to prove $\left(\frac{A_{n+1}}{A_n}\right)^{n+1} \geq \frac{a_n+1}{A_n}$ where $A_n=\frac{a_1+a_2+...+a_n}{n}$ and $a_1, a_2, ...,a_n\in\Bbb{R^+}$

(ii) Use Induction to prove $A_n≥G_n$ where $G_n=\sqrt[n]{a_1a_2...a_n}$

(iii) Hence prove $(a_1+a_2+a_3+...+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}\right)≥n^2$

I've done parts (i) and (ii) but I'm struggling to use the $A_n≥G_n$ to substitute for$\left(\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}\right)$ because if I try to make it one fraction the top becomes unmanageable ($a_2a_3...a_n+a_1a_3...an+...+a_1a_2...a_{n-1}$) and I'm not sure if I'm supposed to expand it use part (ii) and how would I do that anyway. Any help would be greatly appreciated.

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1 Answers1

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Let $v = [ \sqrt{a_1} , \sqrt{a_2} , ...., \sqrt{a_n} ]^T $

and $w = [ \dfrac{1}{\sqrt{a_1}}, \dfrac{1}{\sqrt{a_2}}, ..., \dfrac{1}{\sqrt{a_n}}]^T$

The by Cauchy-Schwartz inequality,

$ v \cdot w \le | v | | w | $

Thus

$ n \le \sqrt{ a_1 + a_2 + ... + a_n } \sqrt{ \dfrac{1}{a_1} + \dfrac{1}{a_2} + ...+\dfrac{1}{a_n} } $

From which,

$\left( a_1 + a_2 + ... + a_n \right) \left( \dfrac{1}{a_1} + \dfrac{1}{a_2} + ...+\dfrac{1}{a_n} \right) \ge n^2 $

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    Please note that 1) the author specifically asked for a proof using the AM-GM inequality, and 2) this has been asked and answered several times before. – Martin R Nov 12 '21 at 22:56