Let $n$ be postive integers. Show that $$\sum_{i=1}^{n}\sum_{j=1}^{n}[i,j]\le\dfrac{n^3}{5}(n+4)\,,$$ where $[a,b]$ denote the least common multiple of $a$ and $b$.
$S_1=1=\dfrac{1^3}{5}(4+1)=1$
Assume that $n>2$ is an integer such that $$S_{n-1}\leq \dfrac{(n-1)^3}{5}(n+3),$$ Then, $$S_{n}-S_{n-1}=n+2\,\sum_{k=1}^{n-1}\,\text{lcm}(k,n)\,.$$
On
If $gcd(i,j)=d$, we have $i=d\,k,\quad j=d\,l,\quad gcd(k,l)=1$, and $lcm(i,j)=d\,k\,l,$ so $$S_n=\sum^n_{d=1}\,d\,\sum_{k,l\le n/d,\,gcd(k,l)=1}k\,l.$$ Let $$T_n=\sum_{k,l\le n,\,gcd(k,l)=1}\,k\,l.$$ Now $$T_n=\sum_{k,l\le n}\,\sum_{d|gcd(k,l)}\,\mu(d)\,k\,l=\sum^n_{d=1}\,\mu(d)\,d^2\,\sum_{k',l'\le n/d}\,k'\,l',$$ and $$\sum_{k',l'\le n/d}\,k'\,l'=\left(\frac{\lfloor n/d\rfloor\,(\lfloor n/d\rfloor+1)}2\right)^2=h(\lfloor n/d\rfloor)$$ with $$h(x)=\frac{x^2\,(x+1)^2}4.$$ Obviously, $$\sum^n_{d=1}\,\mu(d)\,d^2\,h(n/d)=\frac1{4\,\zeta(2)}\,n^4+O(n^3\,\log(n)),$$ and the error introduced by replacing $h(\lfloor n/d\rfloor)$ by $h(n/d)$ is $O(n^3\,\log(n))$ as well. This means $$S_n = \frac{\zeta(3)}{4\,\zeta(2)}\,n^4+O(n^3\,\log(n)),$$ in good agreement with numerical results. It's possible (though tedious) to make the error terms more explicit, so we would have an explict $n$ where the inequality is valid, and we could show it for smaller $n$ by numerical computation. That's the usual, technical and not all too pretty math. If there's a more elegant way to show the inequality, it should be rather smart, since $\frac{\zeta(3)}{4\,\zeta(2)}$ is pretty close to $1/5$.
Let $S_n = \sum \limits_{i=1}^{n} \sum \limits_{j=1}^{n} [i,j]$ and we want to prove that $S_n \leq \frac{n^3}{5}(n+4)$ by induction, its true for $n\leq 100$, we want to prove that $S_{n+6} \leq \frac{(n+6)^3}{5}(n+10)$.
$S_{n+6}=\sum \limits_{i=1}^{n+6} (\sum \limits_{j=1}^{n} [i,j]+[i,n+1]+\cdots+[i,n+6]) =\sum \limits_{i=1}^{n+6} \sum \limits_{j=1}^{n} [i,j]+\sum \limits_{i=1}^{n+6} ([i,n+1]+\cdots+[i,n+6]) = \sum \limits_{i=1}^{n} \sum \limits_{j=1}^{n} [i,j]+\sum \limits_{i=1}^{n+6} ([i,n+1]+\cdots+[i,n+6]) +\sum \limits_{j=1}^{n}([n+1,j]+\cdots+[n+6,j])=S_n +2 \sum \limits_{j=1}^{n}([n+1,j]+\cdots+[n+6,j]) + [n+1,n+1]+[n+1,n+2]+[n+1,n+3]+\cdots+[n+6,n+4]+[n+6,n+5]+[n+6,n+6]$ and so if we put $T_n = \sum \limits_{j=1}^{n}([n+1,j]+\cdots+[n+6,j])$ we get that $S_{n+6} \leq S_n + 2T_n +30n^2+216n+371$
So we need to prove that $S_n+2T_n+30n^2+216n+371 \leq \frac{(n+6)^3}{5}(n+10)$ by the hypothesis we only need to prove that $ \frac{n^3}{5}(n+4)+2T_n +30n^2+216n+371 \leq \frac{(n+6)^3}{5}(n+10)$ or equivalently $T_n \leq 2.4n^3+13.8n^2+21.6n+6.1$ , the proof goes by checking all possible value of $n \mod 60$, but i will only do it for $n=0 \mod 60$, $n = 56\mod 60$ which is the one that makes $T_n$ as big as possible(OP can check for himself).
Assume $n = 0\mod 60$ so $n+2 = 0 \mod 2$ and $n+3 = 0 \mod 3$ and $n+4 = 0 \mod 4$ and $n+5 = 0 \mod 5$ and $n+6=0 \mod 6$ , and so :
$T_n \leq (n+1)\sum \limits_{j=0}^{n} j + (n+2) \sum \limits_{j=0}^{\frac{n}{2}} (2j+1+\frac{2j+2}{2}) +(n+3)\sum \limits_{j=0}^{\frac{n}{3}}(3j+1+3j+2+\frac{3j+3}{3})+(n+4)\sum \limits_{j=0}^{\frac{n}{4}} (4j+1+\frac{4j+2}{2}+4j+3+\frac{4j+4}{4}) +(n+5) \sum \limits_{j=0}^{\frac{n}{5}} (5j+1+5j+2+5j+3+5j+4+\frac{5j+5}{5})+(n+6)\sum \limits_{j=0}^{\frac{n}{6}} (6j+1+\frac{6j+2}{2}+\frac{6j+3}{3}+\frac{6j+4}{2}+6j+5+\frac{6j+6}{6}) = \frac{16549 n^3}{7200}+\frac{341 n^2}{15}+\frac{389 n}{4}+155 \leq 2.4n^3+13.8n^2+21.6n+6.1 $
for all $n \geq 95.92$, checking finite numbers for finite number of cases $0 \mod 60,1 \mod 60 ,\cdots,59 \mod 60$ conclude the proof.
I tried to give a simple proof starting the way OP started his attempt.