Htykuut

Have a short question to this equation.
I have to proof it per induction.
$$sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k} geq n^2 = ncdot n$$

...

so:

For 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 proof that:
$$sum_{k=1}^{n+1} a_kcdotsum_{k=1}^{n+1} dfrac{1}{a_k} geq (n+1)^2 = n^2+2n+1$$

At the end of my proof there are following lines:

$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) + left(dfrac{1}{a_{n+1}}cdot sum_{k=1}^n a_kright) +left((a_{n+1}) cdot sum_{k=1}^n dfrac{1}{a_k}right) + 1 $$

Now we can see that this part
$$left(sum_{k=1}^n a_kcdotsum_{k=1}^n dfrac{1}{a_k}right) +1 geq n^2+1$$

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 proofed.

Is this right, and finally the end? :)

Hint: In the inductive step, observe that $displaystyle sum_{k=1}^nleft(dfrac{a_{n+1}}{a_k}+dfrac{a_k}{a_{n+1}}right)ge 2n$ 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 $+1ge n^2+2n+1=(n+1)^2=RHS$ ,completing the proof.

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