How prove this inequality $\sum_{i=1}^{n}\sum_{j=1}^{n}\text{lcm}(i,j)\le\frac{n^3}{5}(n+4)$? 
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,
$$S_{n}-S_{n-1}=n+2\,\sum_{k=1}^{n-1}\,\text{lcm}(k,n)\,.$$
 A: 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 $1/5$.
