How can I prove $(a+b+c)!>a!b!c!$ In fact, I couldn't prove the inequality because I don't know which method is used for this.
The condition for this inequality is  $$(a+b+c)>1$$  

An adaptation of this answer. If you have $a$ adults, $b$ boys, and $c$ girls, then $(a+b+c)!$ counts the number of ways you can line everyone up, while $a!b!c!$ counts the number of ways you can do so with the additional requirement that all adults come before all children, and all boys before all girls. Then $(a+b+c)!\geq a!b!c!$ always holds, while $(a+b+c)!>a!b!c!$ holds as soon as at least two categories are non-empty, since then there will be some line-up that violates the additional requirement.
