Sequence proof (by induction, presumably) giving me trouble.

Question

Let $a_1,...,a_n$ be a sequence of positive numbers. Show that

$$(a_1+a_2+\cdots+a_n)\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)\geq n^2$$

Hint: Use the fact that for $x>0$ we have $x+(1/x)\geq 2$.

My idea is to use induction. But I can't seem to make it work. On the other hand, I don't see how the hint is going to take me where I need to go.

Are you familiar with the Arithmetic Mean Geometric Mean Inequality (AM/GM)? — André Nicolas, Aug 28 '14 at 17:50
I mean the generalization $\frac{x_1+\cdots+x_n}{n}\ge (x_1x_2\cdots x_n)^{1/n}$. — André Nicolas, Aug 28 '14 at 17:55
No one has mentioned the Cauchy-Schwarz inequality so far. A proof using it is extremely direct, no thinking required at all. — user26486, Aug 29 '14 at 02:59
See also: Is it true that $\forall n \in \Bbb{N} : (\sum_{i=1}^{n} a_{i} ) (\sum_{i=1}^{n} \frac{1}{a_{i}} ) \ge n^2$ , if all $a_{i}$ are positive? and How do I prove this using question using Cauchy-Schwarz inequality? — Martin Sleziak, Jan 21 '19 at 13:56

score 8 · Answer 1 · edited Mar 10 '24 at 00:17

Induction is not necessary. Note that:

\begin{align} (a_1+a_2+\cdots+a_n)&\left(\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right) \\&= \left(\frac{a_1}{a_1} + \frac{a_1}{a_2} + \cdots + \frac{a_1}{a_n} \right)+\cdots + \left(\dfrac{a_n}{a_1}+\dfrac{a_n}{a_2}\cdots+\frac{a_n}{a_n}\right)\\&= \sum_{i=1}^n\sum_{j=1}^n\frac{a_i}{a_j} = \frac{1}{2} \sum_{i=1}^n\sum_{j=1}^n\left(\frac{a_i}{a_j} + \frac{a_j}{a_i}\right) \\ &\stackrel{*}{\geq} \frac{1}{2}\sum_{i=1}^n\sum_{j=1}^n 2 = \frac{1}{2}\sum_{i=1}^n 2n = \frac{1}{2}·2n^2 = n^2\end{align}

Where in $*$ we used the hint $x + \displaystyle\frac{1}{x} \geq 2$ taking $x = \displaystyle\frac{a_i}{a_j}$

change $\frac{a_i}{a_j} + \frac{a_j}{a_i}$ to $\left(\frac{a_i}{a_j} + \frac{a_j}{a_i}\right)$ will look better. — mike, Aug 29 '14 at 01:46

score 5 · Answer 2 · answered Aug 28 '14 at 17:55

5

This is equivalent to:

$$\frac{a_1+a_2+\ldots+a_n}{n}\geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}}$$

Where the LHS is the Arithmetic Mean and the RHS is the Harmonic Mean.

Those are power means defined as:

$$M_p = \left(\frac{1}{n}\sum_{k=1}^na_k^p\right)^{1/p}$$

It is known that $M_p \geq M_q$ iff $p \geq q$.

Note that the Arithmetic Mean is $M_1$ and the Harmonic Mean is $M_{-1}$

answered Aug 28 '14 at 17:55

Darth Geek

12,543

You left to say "You were fast" and upvote :) – pointer Aug 28 '14 at 17:58
Do you know of a method that utilizes the given hint? – nonremovable Aug 28 '14 at 18:00
@user165130 check my other answer ;) – Darth Geek Aug 28 '14 at 18:15
It looks good! Thanks for the help. – nonremovable Aug 28 '14 at 18:21

score 2 · Answer 3 · answered Aug 28 '14 at 18:12

2

Induction works just fine with the hint:

$$(a_1+\ldots+a_n)\left(\frac{1}{a_1}+\ldots+\frac{1}{a_n}\right)=(a_1+\ldots+a_{n-1})\left(\frac{1}{a_1}+\ldots+\frac{1}{a_{n-1}}\right)+\frac{a_1}{a_n}+\frac{a_n}{a_1}+\ldots+\frac{a_{n-1}}{a_n}+\frac{a_n}{a_{n-1}}+1$$

answered Aug 28 '14 at 18:12

WimC

33,414
2
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Ah, I think I see it now. Thank you. – nonremovable Aug 28 '14 at 18:14

score 1 · Answer 4 · answered Aug 28 '14 at 18:14

Clearly is valid for $n=1$. Assume that it holds for $n$ positive numbers. $$\Big((a_1+\cdots+a_n)+a_{n+1}\Big)\Big((\frac{1}{a_1}+\cdots+\frac{1}{a_n})+\frac{1}{a_{n+1}}\Big)=(a_1+\cdots+a_n)(\frac{1}{a_1}+\cdots+\frac{1}{a_n})$$ $$+(\frac{a_1}{a_{n+1}}+\cdots\frac{a_n}{a_{n+1}})+(\frac{a_{n+1}}{a_{1}}+\cdots\frac{a_{n+1}}{a_{n}})+1$$ $$\geq n^2+\Big((\frac{a_{n+1}}{a_{1}}+\frac{a_1}{a_{n+1}})+\cdots(\frac{a_{n+1}}{a_{n}}+\frac{a_n}{a_{n+1}})\Big)+1\geq n^2+2n+1=(n+1)^2.$$ Notice that $x+\frac{1}{x}\geq 2$ as long as $x>0.$

Sequence proof (by induction, presumably) giving me trouble.

4 Answers4