If $a>b$, we immediately get $3a \geqslant 2b+2$. In the case $a=b$, the required inequality is equivalent to $a \geqslant 2$, which can be checked easily since $(a, b)=(1,1)$ does not satisfy $a!+b! \mid a!b!$. We now assume $a > b$. We observe that $c! \mid M$ since $M$ is a product of $c$ consecutive integers. Thus $\operatorname{gcd}(1+M, c!)=1$, which implies
$$1+M \left\lvert\, \frac{a!}{c!}=(c+1)(c+2) \cdots a\right.$$
If $a \leqslant 2c$, then $\frac{a!}{c!}$ is a product of $a-c \leqslant c$ integers not exceeding $a$ whereas $M$ is a product of $c$ integers exceeding $a$. Therefore, $1+M>\frac{a!}{c!}$, which is a contradiction.
It remains to exclude the case $a=2c+1$. Since $a+1=2(c+1)$, we have $c+1 \mid M$. Hence, we can deduce from (1) that $1+M \mid (c+2)(c+3) \cdots a$. Now $(c+2)(c+3) \cdots a$ is a product of $a-c-1=c$ integers not exceeding $a$; thus it is smaller than $1+M$. Again, we arrive at a contradiction.
Comment 1. One may derive a weaker version of (1) and finish the problem as follows. After assuming $a \leqslant 2c+1$, we have $\left\lfloor\frac{a}{2}\right\rfloor \leqslant c$, so $\left.\left\lfloor\frac{a}{2}\right\rfloor!\right\rvert\, M$. Therefore,
$$1+M \left\lvert\,\left(\left\lfloor\frac{a}{2}\right\rfloor+1\right)\left(\left\lfloor\frac{a}{2}\right\rfloor+2\right) \cdots a\right.$$
Observe that $\left(\left\lfloor\frac{a}{2}\right\rfloor+1\right)\left(\left\lfloor\frac{a}{2}\right\rfloor+2\right) \cdots a$ is a product of $\left\lceil\frac{a}{2}\right\rceil$ integers not exceeding $a$. This leads to a contradiction when $a$ is even since $\left\lceil\frac{a}{2}\right\rceil=\frac{a}{2} \leqslant c$ and $M$ is a product of $c$ integers exceeding $a$.
When $a$ is odd, we can further deduce that $1+M \left\lvert\,\left(\frac{a+3}{2}\right)\left(\frac{a+5}{2}\right) \cdots a\right.$ since $\left.\left\lfloor\frac{a}{2}\right\rfloor+1=\frac{a+1}{2} \right\rvert\, a+1$. Now $\left(\frac{a+3}{2}\right)\left(\frac{a+5}{2}\right) \cdots a$ is a product of $\frac{a-1}{2} \leqslant c$ numbers not exceeding $a$, and we get a contradiction.