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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?

Just apply
$$AMgeq GMgeq HM$$
$$Longrightarrow frac{sum_{i=1}^n a_i}{n}geq frac{n}{sum_{i=1}^nfrac{1}{a_i}}$$
Now the result is immediate.

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

Hope it helps:)

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*}

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 $x+yge2sqrt{xy}$, hence $$A ge 2sqrt{(sum_{i=1}^{n} a_i)(sum_{i=1}^{n} frac{1}{a_i})}.$$ So, by induction hypothesis, $A ge 2n$. That's it.

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

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}{ineq 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 + 2binom{n}{2} = n^2\
end{eqnarray*}