Let $n=p\_1^{\alpha\_1}\cdots p\_r^{\alpha\_r}$ be the prime decomposition of the integer $n$. Define $$n' = n \sum\_{i=1}^r \frac{\alpha\_i}{p\_i}\quad\text{and}\quad\Omega(n) = \sum\_{i=1}^r \alpha\_i\quad\text{and}\quad\omega(n) = r.$$ Let $a,b$ be relatively prime, i.e., $\gcd(a,b)=1$, and let $c = a+b$. Suppose that $$\Omega(c) = \min\bigl\{\Omega(a),\Omega(b),\Omega(c)\bigr\}.$$ Is it true that $$\Omega(\gcd(a,a'))+\Omega(\gcd(b,b'))+\Omega(\gcd(c,c')) \le \Omega(ab)-1?$$ From this it would follow that
$$\min\bigl\{\Omega(a),\Omega(b),\Omega(c)\bigr\} \le \omega(abc) - 1.$$


Edit: From the answer given by Kevin Buzzard, one can see, that the first inequality is wrong. It is unclear to me however, if the second inequality is also wrong.



A question about numbers
No. How about $a=9$ and $b=16$? Then $c=25$ so $\Omega(a)=\Omega(c)=2\leq\Omega(b)$, the gcd's are $3,16,5$ so the left hand side is 6 and the right hand side only 5.


Edit: if $a=316$ and $b=27$ then we even have the left hand side being greater than $\Omega(ab)$, answering a question in the comments.


Editedit: if $a=544$ and $b=81$ then indeed your "from this it would follow" inequality is wrong.


