TITLE: $a!b!$ multiple of $a! + b!$ implies $3a\geq 2b + 2$
QUESTION [15 upvotes]: A colleague asked me to show that if $a,b\in\mathbb{N}^*$ are such that $a!b!$  is a multiple of $a! + b!$, then $3a\geq 2b + 2$.
While I could verify this inequality numerically for many solutions $(a,b)$, I did not find anything relevant mathematically. I must say I have never faced such inequalities in arithmetic... 
The problem is symmetric in $a$ and $b$, and actually I could see that the solutions where all such that $3a\geq 2b + 2$ and $3b\geq 2a + 2$.
Here is a plot of the first solutions $(a,b)$ and the bounds:

The solutions can be generated with the Mathematica code:
Table[If[IntegerQ[a!*b!/(a! + b!)], {a, b}, 
 Unevaluated[Sequence[]]], {a, 1, 100}, {b, 2, 100}]

REPLY [4 votes]: 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.