# 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?

$$\sum_{k=1}^n a_k \geq n$$ $$\sum_{k=1}^n \dfrac{1}{a_k} \geq n$$

--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? :)

• Shouldn't $a_n$ be positive? Nov 22, 2018 at 22:14
• I presume you want each $a_i>0$. Nov 22, 2018 at 22:14
• $\sum a_k$ can be any positive number. Nov 22, 2018 at 22:15
• Yeah it can be any positive number. Can i assume in this inequality that the one sum is bigger than n and the other sum is bigger than n? Nov 22, 2018 at 22:18

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.
$$a_n$$ must be always positive. Here is a better way to attain to this:$$\sum_{k=1}^{n}a_k\sum_{l=1}^{n}{1\over a_l}=\sum_{k,l}{a_k\over a_l}=\sum_{kl}{a_k\over a_l}+\sum_{k=l}{a_k\over a_l}=n+\sum_{kalso we know that $${x\over y}+{y\over x}\ge2$$therefore$$n+\sum_{kwhich completes the proof.