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I've been struggling for several hours, trying to prove this horrible inequality:
$(a_1+a_2+dotsb+a_n)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_n}right)geq n^2$.

Where each $a_i$'s are positive and $n$ is a natural number.

First I tried the usual "mathematical induction" method, but it made no avail, since I could not show it would be true if $n=k+1$.

Suppose the inequality holds true when $n=k$, i.e.,

$(a_1+a_2+dotsb+a_k)left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}right)geq n^2$.

This is true if and only if

$(a_1+a_2+dotsb+a_k+a_{k+1})left(frac{1}{a_1}+frac{1}{a_2}+dotsb+frac{1}{a_k}+frac{1}{a_{k+1}}right) -a_{k+1}left(frac{1}{a_1}+dotsb+frac{1}{a_k}right)-frac{1}{a_{k+1}}(a_1+dotsb+a_k)-frac{a_{k+1}}{a_{k+1}} geq n^2$.

And I got stuck here.

The question looks like I have to use AM-GM inequality at some point, but I do not have a clue. Any small hints and clues will be appreciated.

Hint: AM-GM implies
$$
a_1+a_2+cdots +a_nge nsqrt[n]{a_1a_2cdots a_n}
$$ and $$
frac1{a_1}+frac1{a_2}+cdots +frac1{a_n}ge frac{n}{sqrt[n]{a_1a_2cdots a_n}}.
$$

It is AM-HM inequality
$$frac{a_1+a_2+a_3+...+a_n}{n}geq frac{n}{frac{1}{a_1}+frac{1}{a_2}+frac{1}{a_3}+...+frac{1}{a_n}}$$

Here is the proof by induction
that you wanted.

I added a more exact
version of
the identity used
in the proof
at the end.

Let
$s_n
=u_nv_n
$
where
$u_n=sum_{k=1}^n a_k,
v_n= sum_{k=1}^n dfrac1{a_k}
$.
Then,
assuming
$s_n ge n^2$,

$begin{array}\
s_{n+1}
&=u_{n+1}v_{n+1}\
&=(u_n+a_{n+1}) (v_n+dfrac1{a_{n+1}})\
&=u_nv_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&ge n^2+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
end{array}
$

So it is sufficient
to show that
$u_ndfrac1{a_{n+1}}+v_na_{n+1}
ge 2n
$.

By simple algebra,
if $a, b ge 0$ then
$a+b
ge 2sqrt{ab}
$.
(Rewrite as
$(sqrt{a}-sqrt{b})^2ge 0$
or,
as an identity,
$a+b
=2ab+(sqrt{a}-sqrt{b})^2$.)

Therefore

$begin{array}\
u_ndfrac1{a_{n+1}}+v_na_{n+1}
&ge sqrt{(u_ndfrac1{a_{n+1}})(v_na_{n+1})}\
&= sqrt{u_nv_n}\
&=2sqrt{s_n}\
&ge 2sqrt{n^2}
qquadtext{by the induction hypothesis}\
&=2n\
end{array}
$

and we are done.

I find it interesting that
$s_n ge n^2$
is used twice in the
induction step.

Note that,
if we use the identity above,
$a+b
=2ab+(sqrt{a}-sqrt{b})^2$,
we get this:

$begin{array}\
s_{n+1}
&=s_n+u_ndfrac1{a_{n+1}}+a_{n+1}v_n+1\
&=s_n+2sqrt{u_ndfrac1{a_{n+1}}a_{n+1}v_n}+1+(sqrt{u_ndfrac1{a_{n+1}}}-sqrt{a_{n+1}v_n})^2\
&=s_n+2sqrt{s_n}+1+dfrac1{sqrt{a_{n+1}}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&=(sqrt{s_n}+1)^2+dfrac1{sqrt{a_{n+1}}}(sqrt{u_n}-a_{n+1}sqrt{v_n})^2\
&ge(sqrt{s_n}+1)^2\
end{array}
$

with equality
if and only if
$a_{n+1}
=sqrt{dfrac{u_n}{v_n}}
=sqrt{dfrac{sum_{k=1}^n a_k}{sum_{k=1}^n dfrac1{a_k}}}
$.