Proving $\sum\limits_{k=1}^{n+1} a_k\cdot\sum\limits_{k=1}^{n+1} \frac{1}{a_k} \geq (n+1)^2$

Question

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? :)

Answer 1

Hint: In the inductive step, observe that $\displaystyle \sum_{k=1}^n\left(\dfrac{a_{n+1}}{a_k}+\dfrac{a_k}{a_{n+1}}\right)\ge 2n$ by AM-GM inequality, and together with the inductive step we're done.I want to point out that in using your answer $LHS =$ inductive part $+$ the two parts $+1\ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.

Answer 2

$a_n$ 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.