Proving $\sum\limits_{k=1}^{n+1} a_k\cdot\sum\limits_{k=1}^{n+1} \frac{1}{a_k} \geq (n+1)^2$ I have a short question about this equation.
I have to prove it by induction.
$$\sum_{k=1}^n a_k\cdot\sum_{k=1}^n \dfrac{1}{a_k} \geq n^2 = n\cdot n$$
...
so:
By induction proof can I use this? Is the following step correct?
$$\sum_{k=1}^n a_k \geq n$$
$$\sum_{k=1}^n \dfrac{1}{a_k} \geq n$$
Thanks for your answers!
--EDIT--
I should prove that:
$$\sum_{k=1}^{n+1} a_k\cdot\sum_{k=1}^{n+1} \dfrac{1}{a_k} \geq (n+1)^2 = n^2+2n+1$$
At the end of my proof there are the following lines:
$$\left(\sum_{k=1}^n a_k\cdot\sum_{k=1}^n \dfrac{1}{a_k}\right) + \left(\dfrac{1}{a_{n+1}}\cdot \sum_{k=1}^n a_k\right) +\left((a_{n+1}) \cdot \sum_{k=1}^n \dfrac{1}{a_k}\right) + 1 $$
Now we can see that this part
$$\left(\sum_{k=1}^n a_k\cdot\sum_{k=1}^n \dfrac{1}{a_k}\right) +1 \geq n^2+1$$
But what about the other 2 parts?
--- NEW STEPS ---
$$\left(\sum_{k=1}^n \dfrac{a_k}{a_{n+1}}+ \dfrac{a_{n+1}}{a_k}\right)\geq 2n$$
so now I substitute these:
$$\left(\sum_{k=1}^n b_k+ \dfrac{1}{b_k}\right)\geq 2n$$
Induction start:
...
Induction end:
$$\left(\sum_{k=1}^{n+1} b_k+ \dfrac{1}{b_k}\right) = \left(\sum_{k=1}^n b_k+ \dfrac{1}{b_k}\right) + \left(\sum_{k=n+1}^{n+1} b_k+ \dfrac{1}{b_k}\right) $$
$$ \left(\sum_{k=1}^n b_k+ \dfrac{1}{b_k}\right) \geq 2n$$
$$ \left(\sum_{k=n+1}^{n+1} b_k+ \dfrac{1}{b_k}\right) \geq 2$$
So the whole of this:
$$\left(\sum_{k=1}^n \dfrac{a_k}{a_{n+1}}+ \dfrac{a_{n+1}}{a_k}\right)\geq 2n$$
and now the first equation is proved.
Is this right, and finally the end? :)

$a_n$ must be always positive. Here is a better way to attain to this:$$\sum_{k=1}^{n}a_k\sum_{l=1}^{n}{1\over a_l}=\sum_{k,l}{a_k\over a_l}=\sum_{k<l}{a_k\over a_l}+\sum_{k>l}{a_k\over a_l}+\sum_{k=l}{a_k\over a_l}=n+\sum_{k<l}\left({a_k\over a_l}+{a_l\over a_k} \right)$$also we know that $${x\over y}+{y\over x}\ge2$$therefore$$n+\sum_{k<l}\left({a_k\over a_l}+{a_l\over a_k} \right)\ge n+\sum_{k<l}2=n+2\cdot {n^2-n\over 2}=n^2$$which completes the proof.
