$a!b!$ multiple of $a! + b!$ implies $3a\geq 2b + 2$

Assume wlog. that $b\ge a$. Then both $a!b!$ and $a!+b!$ are divisible by $a!$ and $a!+b!\mid a!b!$ implies $1+\frac{b!}{a!}\mid b!$. For $a<k\le b$, $1+\frac{b!}{a!}$ is coprime to $k$. We conclude that $1+\frac{b!}{a!}\mid a!$.

For $1\le k\le b-a$, $k$ has a multiple $qk$ with $a<qk\le b$, hence is again coprime to $1+\frac{b!}{a!}$. We conclude $1+\frac{b!}{a!}\mid \frac{a!}{(b-a) !}$, in particular $$\tag1\frac{b!}{a!}< \frac{a!}{(b-a) !}.$$ By comparing the not cancelled factors in $(1)$ with $a$, we obtain $$ a^{b-a}<a^{2a-b}$$ and hence (for $a>1$) $$\tag2 2b<3a.$$

To improve this, observe that one of $b-a+1, b-a+2$ is not prime (or we are in the trivial case $b=a+1$), hence can be written as product of numbers $<b-a$ and is once again coprime to $1+\frac{b!}{a!}$. This allows us do divide off one more factor on the right of $(1)$, thus leading to $2b<3a-1$ and hence $$ 2b+2\le 3a.$$

This problem is $2015$ IMO Shortlist Number theory N$2$. Two solutions can be found here on page $66$.

Some solutions can be found, e.g., here on Art of Problem Solving.