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This question already has an answer here:

If $\forall i \in \Bbb{N}: a_{i} \in \Bbb{R}^+$ , is it true that $\forall n \in \Bbb{N} : \big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$ ?

I have been able to prove that this holds for $n=1$ , $n=2$, and $n=3$ using the following lemma:

Lemma 1: Let $a,b \in \Bbb{R}^+$. If $ab =1$ then $a+b \ge 2$

For example, the case for $n=3$ can be proven like this:

Let $a,b,c \in \Bbb{R}^+$. Then we have:

$(a+b+c)\big(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\big) = 1 + \frac{a}{b} + \frac{a}{c} + \frac{b}{a} + 1 + \frac{b}{c} + \frac{c}{a} + \frac{c}{b} + 1 $

$= 3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) $

By lemma 1, $\big(\frac{a}{b} + \frac{b}{a}\big) \ge 2$, $ \big(\frac{a}{c} + \frac{c}{a}\big) \ge 2$ and $\big(\frac{b}{c} + \frac{c}{b}\big) \ge 2$ , therefore:

$3 + \big(\frac{a}{b} + \frac{b}{a}\big) + \big(\frac{a}{c} + \frac{c}{a}\big) + \big(\frac{b}{c} + \frac{c}{b}\big) \ge 3 + 2 + 2 +2 = 9 = 3^2 \ \blacksquare $

However I'm not sure the generalized version for all natural $n$ is true. I can't come up with a counterexample and when I try to prove it by induction I get stuck.

Here is my attempt:

Let $P(n)::\big(\sum_{i=1}^{n}a_{i}\big) \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big) \ge n^2$

Base case: $\big(\sum_{i=1}^{1}a_{i}\big) \big(\sum_{i=1}^{1} \frac{1}{a_{i}}\big) = a_{1} \frac{1}{a_{1}} = 1 = 1^2$ , so $P(1)$ is true.

Inductive hypothesis: I assume $P(n)$ is true.

Inductive step:

$$\left(\sum_{i=1}^{n+1}a_{i}\right) \left(\sum_{i=1}^{n+1} \frac{1}{a_{i}}\right) = \left[\left(\sum_{i=1}^{n}a_{i}\right) + a_{n+1}\right] \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$

$$=\left(\sum_{i=1}^{n}a_{i}\right) \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right] + a_{n+1} \left[\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + \frac{1}{a_{n+1}}\right]$$

$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +a_{n+1} \frac{1}{a_{n+1}}$$

$$=\left(\sum_{i=1}^{n}a_{i}\right)\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +\left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) +1 $$

$$\underbrace{\ge}_{IH} n^2 + \left(\sum_{i=1}^{n}a_{i}\right) \frac{1}{a_{n+1}} + a_{n+1} \left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) + 1$$

And here I don't know what to do with the $\big( \sum_{i=1}^{n}a_{i} \big) \frac{1}{a_{n+1}} + a_{n+1} \big(\sum_{i=1}^{n} \frac{1}{a_{i}}\big)$ term.

Is this inequality true? If it is, how can I prove it? If it isn't, can anyone show me a counterexample?

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marked as duplicate by Martin R, Did, José Carlos Santos, Martin Sleziak, RRL Jan 21 at 17:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Just apply $$AM\geq GM\geq HM$$ $$\Longrightarrow \frac{\sum_{i=1}^n a_i}{n}\geq \frac{n}{\sum_{i=1}^n\frac{1}{a_i}}$$ Now the result is immediate.

Here equality holds iff all $a_i's$ are equal.

Hope it helps:)

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HINT:

The Cauchy-Schwarz Inequality \begin{eqnarray*} (x_1^2+ \cdots +x_n^2) (y_1^2+ \cdots +y_n^2) \geq (x_1 y_1+ \cdots +x_n y_n)^2 \end{eqnarray*}

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  • $\begingroup$ Nice answer. I was thinking of another way: Use Jensen’s inequality on $\frac{1}{n}\sum_{i =1}^{n}1/a_i$. $\endgroup$ – Michael Jan 21 at 0:54
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Well you could just use AM-HM, Cauchy-Schwarz, Chebyshev or the rearrangement inequality or the like. However, I suppose the point of this exercise is to prove AM-HM so let's continue with your efforts. To use induction we need to show that $$A = \frac{1}{a_{n+1}} \sum_{i=1}^{n} a_i + a_{n+1}\sum_{i=1}^{n} \frac{1}{a_i} \ge 2n.$$ It's not hard to show that $x+y\ge2\sqrt{xy}$, hence $$A \ge 2\sqrt{(\sum_{i=1}^{n} a_i)(\sum_{i=1}^{n} \frac{1}{a_i})}.$$ So, by induction hypothesis, $A \ge 2n$. That's it.

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Also, you can use AM-GM: $$\sum_{k=1}^na_k\sum_{k=1}^n\frac{1}{a_k}\geq n\sqrt[n]{\prod_{k=1}^na_k}\cdot \frac{n}{\sqrt[n]{\prod\limits_{k=1}^na_k}}=n^2.$$

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Just for completeness:

You may also show the inequality by direct calculation using

  • $x + \frac{1}{x} \geq 2$ for $x > 0$

\begin{eqnarray*} \left(\sum_{i=1}^{n} a_{i} \right) \left(\sum_{j=1}^{n} \frac{1}{a_{j}} \right) & = & \sum_{i,j =1}^n \frac{a_i}{a_j} \\ & = & \sum_{\stackrel{i,j =1}{\color{blue}{i=j}}}^n \frac{a_i}{a_j} + \sum_{\stackrel{i,j =1}{\color{blue}{i\neq j}}}^n \frac{a_i}{a_j} \\ & = & n + \sum_{\stackrel{i,j =1}{\color{blue}{i< j}}}^n \left(\frac{a_i}{a_j} + \frac{a_j}{a_i} \right)\\ & \color{blue}{\geq} & n + \sum_{\stackrel{i,j =1}{\color{blue}{i< j}}}^n 2\\ & = & n + 2\binom{n}{2} = n^2\\ \end{eqnarray*}

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