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While reading online, I encountered this post which the author claims that
beginalign
S(N, 1):=sum_1le i, j le N frac1textlcm(i, j) geq 3H_N-2
endalign
and $S(N, 1) geq H_N^2$ where $H_N$ is the partial harmonic sum.

The prove of the latter inequality is already given in the post. For the former, we see that
beginalign
S(N, 1)
=& 2sum^N_j=2sum^j-1_i=1frac1textlcm(i, j) +H_N\
geq& 2sum^N_j=2 frac1textlcm(1, j)+H_N= 3H_N -2.
endalign

My question is whether there exists $C$, independent of $N$, such that the following bound holds
beginalign
s(N):=sum_N leq i, j leq 2N frac1textlcm(i, j) geq C log N.
endalign
Actually, I know for a fact that this is true by $(1.7)$ in this paper but couldn't prove it.

Another question: How does one show
beginalign
sum_N^2/4< k< N^2/3 (#left qmid kright)^2sim sum_N/3<q, q'<N/2 fracN^2textlcm(q,q').
endalign
I started by
beginalign
sum_N^2/4< k< N^2/3 (#left qmid kright)^2 =& sum_N^2/4< k< N^2/3left( sum_substackN/3<q<N/2\ qmid k1 right)^2\
=& sum_N^2/4< k< N^2/3 sum_substackN/3<q, q'<N/2\ qmid k, q'mid k 1.
endalign
Here I know I should interchange the order of summation, but I don't know how to get the lcm to show up.

Any help/suggestion is highly welcome.

Sketch proof for 1: assume that there is $c>0$ absolute constant s.t for each $1 leq a leq N^frac14$ ($N$ high enough), there are at least $cfracN^2a^2$ pairs $N leq k,l leq 2N$ with $gcd(k,l)=a$; then as $lcm(k,l)gcd(k,l)=kl$,

$s(N):=sum_N leq k, l leq 2N frac1textlcm(k, l) =sum_N leq k, l leq 2N fracgcd(k,l)kl geq frac14N^2sum_N leq k, l leq 2Ngcd(k,l) geq frac14N^2sum_1 leq a leq N^frac14a(cfracN^2a^2) geq fracc16log N $

Let now $1 leq a leq N^frac14$, considering the first number $N leq k(a) < N+a$ which is divisble by $a$, it follows that $k(a)+qa$ is divisble by $a$ and within our range as long as $(q+1)a leq N$, so there are at least $[fracNa-1]$ such, so at least $[fracNa-1]^2$ pairs with gcd divisible by $a$; same reasoning give at most $[fracNa+1]^2$ such pairs, so applying this with $2a,3a...2Na$ (since the gcd is at most $2N$ in our interval) we get that the number of pairs with gcd precisely $a$ is at least: $beginalignfracN^2a^2(1-frac14-frac19-.. )- 2N(log 2N+1) geq fracN^2a^2(1-(fracpi^26-1))-2N(log 2N+1) endalign$ $geq fracN^26a^2-2N(log 2N+1)$, and that is $ geq fracN^212a^2$ for say $N geq 40000$ since $a leq N^frac14, fracN^212a^2 geq fracN^212sqrt N geq 2N(log 2N+1)$, for $N$ large as noted so we are done.

For the second claim, note that switching the sum, each $N^2/4< k< N^2/3$ appears for a pair $(q,q')$ precisely when it is a multiple of the lcm of the pair and it is fairly clear that for any $m geq 1$ in a ~$N^2$ interval there are ~$fracN^2m$ multiples of it, so the claim follows

Edit: as requested a more detailed explanation of the estimate above for the number of pairs with gcd $a$:

because the number of multiples of $a$ between $N, 2N$ is between $fracNa-1, fracNa+1$, this means that we have at least $(fracNa-1)^2=fracN^2a^2-2fracNa+1$ pairs with gcd a multiple of $a$, but if the gcd is not $a$, it means it is $2a, 3a,...2Na$ (this last is very crude and we can do much better but we do not need), but for $2a$ we have at most $(fracN2a+1)^2=fracN^24a^2+2fracN2a+1$ pairs with gcd that (actually less since some will have gcd $4a$ etc but we again use a rough estimate as that is enough); putting things together we get by subtracting all those cases with gcd $2a,3a...2Na$, the main term is $fracN^2a^2(1-frac14-frac19-.. )$ which we estimate by $fracN^26a^2$ with a simple computation; the remainder is $(-2fracNa+1)+(-2fracN2a-1)+(-2fracN3a-1)+...$ (at most $2N$ terms) and that clearly estimates by $-2Nlog 2N - 2N$, so we get the estimate above and use that $a leq N^frac14$ to conclude as noted for large enough $N$ since obviously $fracN^212a^2 geq fracN^212sqrt N$ will dominate $2Nlog 2N + 2N$, allowing us to keep another $fracN^212a^2$ from the main term